Math / Algebra

QuestionУпражнения
1. Представить в виде многочлена:
1. $\left(\frac{1}{5}-m\right)^{2}$
2. $\left(2+\frac{1}{8} x\right)^{2}$
3. $\left(-7 y+\frac{1}{7} x\right)^{2}$
4. $\left(x-\frac{3}{4}\right)\left(\frac{3}{4}+x\right)$
5. $(4 x+2 y)^{3}$ II. Разложить на множители:
1. $9 x^{2}-6 x+1$
2. $16 m^{2}+24 m n+9 n^{2}$
3. $0,09 m^{2}-64 n^{2}$
4. $7,29 x^{6}-7,84 y^{6}$
5. $27 a^{3}-64 b^{3}$
6. $(2-x)^{2}-(3 x+5)^{2}$
7. $(3 m+5)^{2}-64$
8. $x^{2}-2 x-3$
9. $7 x^{2}-5 x+3$ III. Сократить дробь
1. $\frac{4 a-4 b}{a^{2}-b^{2}}$
2. $\frac{14 c-8 b}{49 c^{2}-16 b^{2}}$ $\frac{4 x^{2}+20 x+25}{4 x^{2}-25}$ IV. Упростить выражение
1. $\sqrt[3]{2 a b^{2}} \cdot \sqrt[3]{4 a^{2} b}$
2. $\sqrt[4]{\frac{a b}{c}} \cdot \sqrt[4]{\frac{a^{3} c}{b}}$
3. $\frac{\sqrt[5]{a^{6} \cdot b^{7}}}{\sqrt[5]{a \cdot b^{2}}}$
4. $\frac{\sqrt[3]{\frac{3 x}{y^{2}}}}{\sqrt[3]{\frac{y}{9 x^{2}}}}$
5. $\left(\sqrt[6]{x^{4}}\right)^{-3}$
6. $\sqrt{\sqrt[3]{729}}$
7. $\sqrt[3]{\sqrt[3]{9}} \cdot \sqrt[9]{3^{7}}$
8. $\left(\sqrt{\sqrt[3]{a^{2} b}}\right)^{6}$
9. $\left(x^{3 / 4}\right)^{4 / 5}$
10. $\frac{\left(\mathrm{a}^{-3}\right)^{-2} \cdot\left(\mathrm{a}^{3}\right)^{-3}}{\left(\mathrm{a}^{-1}\right)^{-2}:\left(\mathrm{a}^{2}\right)^{-4}}$
11. $\left(\frac{2}{5} a^{-3} b^{2}\right)^{-2}:\left(3 \frac{1}{3} a^{-4} b^{3}\right)^{2}$

Studdy Solution

STEP 1

What is this asking? We're going to have a blast simplifying and factoring some awesome expressions, then we'll conquer some fraction and radical simplification! Watch out! Remember your exponent rules and factoring formulas, and be super careful with signs!

STEP 2

1. Simplify the expression $\left(\frac{1}{5}-m\right)^{2}$
2. Factor the expression $9x^2 - 6x + 1$
3. Simplify the fraction $\frac{4a-4b}{a^2 - b^2}$
4. Simplify the expression $\sqrt[3]{2ab^2} \cdot \sqrt[3]{4a^2b}$

STEP 3

Remember $(a-b)^2 = a^2 - 2ab + b^2$ .
Let $a = \frac{1}{5}$ and $b = m$ .
So, we have: $\left(\frac{1}{5}-m\right)^{2} = \left(\frac{1}{5}\right)^{2} - 2 \cdot \frac{1}{5} \cdot m + m^{2}$

STEP 4

$\left(\frac{1}{5}\right)^{2} = \frac{1^2}{5^2} = \frac{1}{25}$ $-2 \cdot \frac{1}{5} \cdot m = -\frac{2}{5}m$ Putting it all together: $\frac{1}{25} - \frac{2}{5}m + m^2$

STEP 5

This looks like a perfect square trinomial!
The pattern is $a^2 - 2ab + b^2 = (a-b)^2$ .

STEP 6

Here, $a^2 = 9x^2$ , so $a=3x$ .
Also, $b^2 = 1$ , so $b=1$ .
Let's check the middle term: $-2ab = -2(3x)(1) = -6x$ .
Perfect!

STEP 7

So, $9x^2 - 6x + 1 = (3x-1)^2$ .

STEP 8

We can factor out a 4 from the numerator: $4a - 4b = 4(a-b)$ .

STEP 9

The denominator is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$ .

STEP 10

Now we have: $\frac{4(a-b)}{(a-b)(a+b)}$ We can divide both the numerator and denominator by $(a-b)$ , effectively multiplying by one in the form of $\frac{1/(a-b)}{1/(a-b)}$ .
This simplifies to: $\frac{4}{a+b}$

STEP 11

Since the roots are the same (cube roots), we can multiply the radicands: $\sqrt[3]{2ab^2 \cdot 4a^2b} = \sqrt[3]{8a^3b^3}$

STEP 12

$\sqrt[3]{8a^3b^3} = \sqrt[3]{(2ab)^3}$

STEP 13

$\sqrt[3]{(2ab)^3} = 2ab$

STEP 14

1. $\left(\frac{1}{5}-m\right)^{2} = \frac{1}{25} - \frac{2}{5}m + m^2$
2. $9x^2 - 6x + 1 = (3x-1)^2$
3. $\frac{4a-4b}{a^2 - b^2} = \frac{4}{a+b}$
4. $\sqrt[3]{2ab^2} \cdot \sqrt[3]{4a^2b} = 2ab$

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Studdy Solution

STEP 1

What is this asking? We're going to have a blast simplifying and factoring some awesome expressions, then we'll conquer some fraction and radical simplification! Watch out! Remember your exponent rules and factoring formulas, and be super careful with signs!

STEP 2

1. Simplify the expression $\left(\frac{1}{5}-m\right)^{2}$
2. Factor the expression $9x^2 - 6x + 1$
3. Simplify the fraction $\frac{4a-4b}{a^2 - b^2}$
4. Simplify the expression $\sqrt[3]{2ab^2} \cdot \sqrt[3]{4a^2b}$

STEP 3

Remember $(a-b)^2 = a^2 - 2ab + b^2$ .
Let $a = \frac{1}{5}$ and $b = m$ .
So, we have: $\left(\frac{1}{5}-m\right)^{2} = \left(\frac{1}{5}\right)^{2} - 2 \cdot \frac{1}{5} \cdot m + m^{2}$

STEP 4

$\left(\frac{1}{5}\right)^{2} = \frac{1^2}{5^2} = \frac{1}{25}$ $-2 \cdot \frac{1}{5} \cdot m = -\frac{2}{5}m$ Putting it all together: $\frac{1}{25} - \frac{2}{5}m + m^2$

STEP 5

This looks like a perfect square trinomial!
The pattern is $a^2 - 2ab + b^2 = (a-b)^2$ .

STEP 6

Here, $a^2 = 9x^2$ , so $a=3x$ .
Also, $b^2 = 1$ , so $b=1$ .
Let's check the middle term: $-2ab = -2(3x)(1) = -6x$ .
Perfect!

STEP 7

So, $9x^2 - 6x + 1 = (3x-1)^2$ .

STEP 8

We can factor out a 4 from the numerator: $4a - 4b = 4(a-b)$ .

STEP 9

The denominator is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$ .

STEP 10

Now we have: $\frac{4(a-b)}{(a-b)(a+b)}$ We can divide both the numerator and denominator by $(a-b)$ , effectively multiplying by one in the form of $\frac{1/(a-b)}{1/(a-b)}$ .
This simplifies to: $\frac{4}{a+b}$

STEP 11

Since the roots are the same (cube roots), we can multiply the radicands: $\sqrt[3]{2ab^2 \cdot 4a^2b} = \sqrt[3]{8a^3b^3}$

STEP 12

$\sqrt[3]{8a^3b^3} = \sqrt[3]{(2ab)^3}$

STEP 13

$\sqrt[3]{(2ab)^3} = 2ab$

STEP 14

1. $\left(\frac{1}{5}-m\right)^{2} = \frac{1}{25} - \frac{2}{5}m + m^2$
2. $9x^2 - 6x + 1 = (3x-1)^2$
3. $\frac{4a-4b}{a^2 - b^2} = \frac{4}{a+b}$
4. $\sqrt[3]{2ab^2} \cdot \sqrt[3]{4a^2b} = 2ab$

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Studdy Solution

STEP 1

What is this asking? We're going to have a blast simplifying and factoring some awesome expressions, then we'll conquer some fraction and radical simplification! Watch out! Remember your exponent rules and factoring formulas, and be super careful with signs!

STEP 2

STEP 3

STEP 4

(15)2=1252=125\left(\frac{1}{5}\right)^{2} = \frac{1^2}{5^2} = \frac{1}{25}(51​)2=5212​=251​ −2⋅15⋅m=−25m-2 \cdot \frac{1}{5} \cdot m = -\frac{2}{5}m−2⋅51​⋅m=−52​m Putting it all together: 125−25m+m2\frac{1}{25} - \frac{2}{5}m + m^2251​−52​m+m2

STEP 5

This looks like a perfect square trinomial!The pattern is a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a-b)^2a2−2ab+b2=(a−b)2.

STEP 6

Here, a2=9x2a^2 = 9x^2a2=9x2, so a=3xa=3xa=3x.Also, b2=1b^2 = 1b2=1, so b=1b=1b=1.Let's check the middle term: −2ab=−2(3x)(1)=−6x-2ab = -2(3x)(1) = -6x−2ab=−2(3x)(1)=−6x.Perfect!

STEP 7

So, 9x2−6x+1=(3x−1)29x^2 - 6x + 1 = (3x-1)^29x2−6x+1=(3x−1)2.

STEP 8

We can factor out a **4** from the numerator: 4a−4b=4(a−b)4a - 4b = 4(a-b)4a−4b=4(a−b).

STEP 9

The denominator is a difference of squares: a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b).

STEP 10

STEP 11

Since the roots are the same (cube roots), we can multiply the radicands: 2ab2⋅4a2b3=8a3b33\sqrt[3]{2ab^2 \cdot 4a^2b} = \sqrt[3]{8a^3b^3}32ab2⋅4a2b​=38a3b3​

STEP 12

8a3b33=(2ab)33\sqrt[3]{8a^3b^3} = \sqrt[3]{(2ab)^3}38a3b3​=3(2ab)3​

STEP 13

(2ab)33=2ab\sqrt[3]{(2ab)^3} = 2ab3(2ab)3​=2ab

STEP 14

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$\left(\frac{1}{5}\right)^{2} = \frac{1^2}{5^2} = \frac{1}{25}$ $-2 \cdot \frac{1}{5} \cdot m = -\frac{2}{5}m$ Putting it all together: $\frac{1}{25} - \frac{2}{5}m + m^2$

This looks like a perfect square trinomial!
The pattern is $a^2 - 2ab + b^2 = (a-b)^2$ .

Here, $a^2 = 9x^2$ , so $a=3x$ .
Also, $b^2 = 1$ , so $b=1$ .
Let's check the middle term: $-2ab = -2(3x)(1) = -6x$ .
Perfect!

So, $9x^2 - 6x + 1 = (3x-1)^2$ .

We can factor out a 4 from the numerator: $4a - 4b = 4(a-b)$ .

The denominator is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$ .

Since the roots are the same (cube roots), we can multiply the radicands: $\sqrt[3]{2ab^2 \cdot 4a^2b} = \sqrt[3]{8a^3b^3}$

$\sqrt[3]{8a^3b^3} = \sqrt[3]{(2ab)^3}$

$\sqrt[3]{(2ab)^3} = 2ab$