@@ đc đưa vào câu hoi hay lun ak

:D ko bt khó không nhưng nhì so qua thì áp dụng

|a|+|b|>=|a+b|

|a|=|-a|

Theo đầu bài ta có:
\(\frac{1}{5}\cdot a+2+\frac{1}{2}\cdot a+7=a\)
\(\Rightarrow2+7=a-\frac{1}{2}\cdot a-\frac{1}{5}\cdot a\)
\(\Rightarrow a\cdot\frac{3}{10}=9\)
\(\Rightarrow a=30\)

\(\frac{1}{5}a+2+\frac{1}{2}a+7=a\left(\frac{1}{5}+\frac{1}{2}\right)+2+7=\frac{7}{10}a+10=\frac{7a}{10}+10\)

\(f\left(x\right)=\left(x-1\right).g\left(x\right)\)

\(\Rightarrow3x^3-2x^2+x+5=\left(x-1\right)\left(3x^2+ax+b\right)\)

\(\Rightarrow3x^3-2x^2+x+5=3x^3+ax^2+bx-3x^2-ax-b\)

\(\Rightarrow-2x^2+x+5=x^2\left(a-3\right)+x\left(b-a\right)-b\)

-Bạn kiểm tra lại đề.

ban lam sai roi 3x^3 + ax^2 phai =x^2.(3+a)

Ta thấy:

Vì a.b= 36 và a < 4

\(\Rightarrow a\inƯC\left\{36\right\}\left(a< 4\right)\)

\(\Rightarrow a\in1;2;3\) và \(b\in12;18;36\)

\(a\in\left\{9;13;18;36\right\}\) và \(b\in\left\{1;2;3;4\right\}\)
 

\(5-\left(6-x\right)=4\left(3-2x\right)\)

\(5-6+x=12-8x\)

\(-1+x=12-8x\)

\(x-1=12-8x\)

\(12+1=8x+1\)

\(8x=13-1\)

\(x=12:8\)

\(x=\dfrac{12}{8}=\dfrac{3}{2}\)

\(PT\Leftrightarrow5-6+x=12-8x\)

\(\Leftrightarrow9x=13\)

\(\Leftrightarrow x=\dfrac{13}{9}\)

Vậy: \(S=\left\{\dfrac{13}{9}\right\}\)

(2x-14).(3x+42)=0

Vậy 2x-14=0 hoặc 3x+42=0

- Nếu 2x-14=0

         2x=0+14

         2x=14

         x=14:2

         x=7

- Nếu 3x+42=0

        3x=0-42

        3x=-42

        x=-42:3

        x=-14

Vậy x=7 hoặc -14

A.B =0 => A =0 hoặc B =0

=> 2x - 14 =0 => x =7

hoặc 3x + 42 =0 => x = -14

a) Ta có: \(6x^4-9x^3\)

\(=3x^3\cdot2x-3x^3\cdot3\)

\(=3x^3\left(2x-3\right)\)

b) Ta có: \(x^2y^2z+xy^2z^2+x^2yz^2\)

\(=xyz\cdot\left(xy+yz+xz\right)\)

c) Ta có: \(2x\left(x+3\right)+2\left(x+3\right)\)

\(=2\cdot\left(x+3\right)\cdot x+2\cdot\left(x+3\right)\cdot1\)

\(=2\left(x+3\right)\left(x+1\right)\)

d) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)

\(=\left(x+5\right)\left(x+5-3\right)\)

\(=\left(x+5\right)\left(x+2\right)\)

e) Ta có: \(2x\left(x-3\right)-\left(x-3\right)^2\)

\(=\left(x-3\right)\left(2x-x+3\right)\)

\(=\left(x-3\right)\left(x+3\right)\)

a, 6x⁴ - 9x³ = 3x³ (2x-3x) = 3x³ (-x) = -3x⁴

b, x²y²z + xy²z² + x²yz² = xyz (xy+yz+xz)

c, 2x (x+3) + 2 (x+3) = (x+3) (2x+2) = (x+3) 2 (x+1)

d, (x+5)² - 3 (x+5) = (x+5) (x+5-3) = (x+5) (x+2)

e, 2x (x-3) - (x-3)² = (x-3) [2x-(x-3)] = (x-3) (2x-x+3) = (x-3) (x+3) = x² - 9

Tự làm á! Đúng sai thì chịu