a: =3(x-y)(x+y)

\(a,=3\left(x-y\right)\left(x+y\right)\\ b,=2x\left(x^2-25\right)=2x\left(x-5\right)\left(x+5\right)\\ c,=5x\left(x^2-2x+1\right)=5x\left(x-1\right)^2\\ d,=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+y+2\right)\\ e,=x\left(y-3\right)+y\left(y-3\right)=\left(y-3\right)\left(x+y\right)\\ f,=\left(x+2\right)^2-16y^2=\left(x-4y+2\right)\left(x+4y+2\right)\)

1: \(=\left(y-1\right)^2\)

2: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

3: =(1-2x)(1+2x)

\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

5: \(=\left(x+3\right)^3\)

6: \(=\left(2x-y\right)^3\)

cảm ơn vị cứu tinh

a) 

<=> \(3x-12x^2+12x^2-6x=9\)

<=> \(-3x=9\)

<=> \(x=-3\)

b)

<=> \(6x-24x^2-12x+24x^2=6\)

<=> \(-6x=6\)

<=> \(x=-1\)

c) 

<=> \(6x-4-3x+6=1\)

<=> \(3x+2=1\)

<=> \(x=-\frac{1}{3}\)

d) 

<=> \(9-6x^2+6x^2-3x=9\)

<=> \(-3x=0\)

<=> \(x=0\)

e) KO HIỂU ĐỀ

f) 

<=> \(4x^2-8x+3-\left(4x^2+9x+2\right)=8\)

<=> \(-17x+1=8\)

<=> \(x=-\frac{7}{17}\)

g) 

<=> \(-6x^2+x+1+6x^2-3x=9\)

<=> \(-2x=8\)

<=> \(x=-4\)

h)

<=> \(x^2-x+2x^2+5x-3=4\)

<=> \(3x^2+4x=7\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{7}{3}\end{cases}}\)

a. \(3x\left(1-4x\right)+6x\left(2x-1\right)=9\)

\(\Rightarrow3x-12x^2+12x^2-6x=9\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=-3\)

b. \(3x\left(2-8x\right)-12x\left(1-2x\right)=6\)

\(\Rightarrow6x-24x^2-12x+24x^2=6\)

\(\Rightarrow-6x=6\)

\(\Rightarrow x=-1\)

c. \(2\left(3x-2\right)-3\left(x-2\right)=1\)

\(\Rightarrow6x-4-3x+6=1\)

\(\Rightarrow3x+2=1\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

11)\(x^3+8x^2+5x+a=x\left(x^2+3x+b\right)+5\left(x^2+3x+b\right)-bx-10x+5b+a=\left(x^2+3x+b\right)\left(x+5\right)-bx-10x+5b+a⋮\left(x^2+3x+b\right)\)

\(\Rightarrow\left\{{}\begin{matrix}-bx-10x=0\\5b+a=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=-10\\a=50\end{matrix}\right.\)

mình không thi  đâu

mk chịu vì ít tuổi hơn

làm hết hồn

Em tách ra mỗi lần hỏi đăng 1-3 bài thôi để nhận hỗ trợ sớm nhất nha em!

Bài 1 

a) 2x+6=0

<=> 2x=-6

x=-3

b) 15 -7x=9-3x

<=> -4x=-6

<=>x=\(\dfrac{3}{2}\)

c) 2(x+1)=5x-7

<=>2x+2=5x-7

<=>-3x=-9x

<=>x=3

Câu 1:

a)\(\left(x+2\right)^2-x\left(x+2\right)=0\Leftrightarrow2\left(x+2\right)=0\Leftrightarrow x=-2\)

b) \(\dfrac{2x+7}{3}-\dfrac{x-2}{4}=2\Leftrightarrow\dfrac{8x+28}{12}-\dfrac{3x-6}{12}=2\)

\(\Leftrightarrow\dfrac{5x+34}{12}=2\Leftrightarrow5x+34=24\Leftrightarrow5x=-10\Leftrightarrow x=-2\)

c)\(\left|x+5\right|=3x+1\Leftrightarrow\left\{{}\begin{matrix}x+5=3x+1\\x+5=-3x-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)

Bài 1: 

b) Ta có: \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)

\(\Leftrightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)

\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)

\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

mà \(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\ne0\)

nên x-89=0

hay x=89

Vậy: S={89}

Bài 1:

a)ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

Ta có: \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}\)

Suy ra: \(x^2+x+x^2-3x-4x=0\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhân\right)\\x=6\left(nhận\right)\end{matrix}\right.\)

Vậy: S={0;6}