b: Ta có: \(B=x^2\left(11x-2\right)+x^2\left(x-1\right)-3x\left(4x^2-x-2\right)\)

\(=11x^3-2x^2+x^3-x^2-12x^3+3x^2+6x\)

\(=6x\)

a: ĐKXĐ: x>=9/4

\(\sqrt{4x-9}=2x-5\)

=>\(\begin{cases}\left(2x-5\right)^2=4x-9\\ 2x-5\ge0\end{cases}\Rightarrow\begin{cases}4x^2-20x+25-4x+9=0\\ x\ge\frac52\end{cases}\)

=>\(\begin{cases}4x^2-24x+34=0\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}2x^2-12x+17=0\\ x\ge\frac52\end{cases}\)

=>\(\begin{cases}x^2-6x+\frac{17}{2}=0\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}x^2-6x+9-\frac12=0\\ x\ge\frac52\end{cases}\)

=>\(\begin{cases}\left(x-3\right)^2=\frac12\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}x-3\in\left\lbrace\frac{\sqrt2}{2};-\frac{\sqrt2}{2}\right\rbrace\\ x\ge\frac52\end{cases}\)

=>\(x=3+\frac{\sqrt2}{2}=\frac{6+\sqrt2}{2}\)

b: ĐKXĐ: \(x^2-7x+10\ge0\)

=>(x-5)(x-2)>=0

=>x>=5 hoặc x<=2

\(\sqrt{x^2-7x+10}=3x-1\)

=>\(\begin{cases}3x-1\ge0\\ \left(3x-1\right)^2=x^2-7x+10\end{cases}\Rightarrow\begin{cases}9x^2-6x+1-x^2+7x-10=0\\ x\ge\frac13\end{cases}\)

=>\(\begin{cases}8x^2+x-9=0\\ x\ge\frac13\end{cases}\Rightarrow\begin{cases}8x^2+9x-8x-9=0\\ x\ge\frac13\end{cases}\)

=>\(\begin{cases}\left(x+1\right)\left(8x-9\right)=0\\ x\ge\frac13\end{cases}\Rightarrow x=\frac98\) (nhận)

d: |3x-1|=x+3

=>\(\begin{cases}x+3\ge0\\ \left(3x-1\right)^2=\left(x+3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-3\\ \left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-3\\ \left(2x-4\right)\left(4x+2\right\rbrace\end{cases}\Rightarrow x\in\left\lbrace2;-\frac12\right\rbrace\)

e: |x+2|=|6-3x|

=>|3x-6|=|x+2|

=>3x-6=x+2 hoặc 3x-6=-x-2

=>2x=8 hoặc 4x=4

=>x=4 hoặc x=1

a) Ta có: \(x^2-8x+7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

b) Ta có: \(x^2+x-20=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)

c) Ta có: \(3x^2+4x-4=0\)

\(\Leftrightarrow3x^2+6x-2x-4=0\)

\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) Ta có: \(3x^2-4x-7=0\)

\(\Leftrightarrow3x^2-7x+3x-7=0\)

\(\Leftrightarrow\left(3x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-1\end{matrix}\right.\)

e) Ta có: \(5x^2-16x+3=0\)

\(\Leftrightarrow5x^2-15x-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

f) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

a)

\(x^2-8x+7=0\text{⇔}\text{⇔}x^2-7x-x-7=\left(x-7\right)\left(x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

Vậy nghiệm của đa thức : \(S=\left\{1;7\right\}\)

c)

\(3x^2+4x-4=0\text{⇔}3x^2+6x-2x-4=\left(3x-2\right)\left(x+2\right)=0\text{⇔}\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

Vậy nghiệm của đa thức : \(S=\left\{\dfrac{2}{3};-2\right\}\)

b)

\(x^2+x-20=0⇔\left(x-4\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

d)

\(3x^2-4x-7=0\text{⇔}\left(3x-7\right)\left(x+1\right)=0\text{⇔}\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)

e)

\(5x^2-16x+3\text{⇔}\left(x-3\right)\left(5x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

f)

\(x^2+3x-10=0\text{⇔}\left(x-2\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

\(\)

1,\(=4x\left(x-\dfrac{3}{2}\right)\)
2,\(=-7y^3\left[2x^2y\left(2y+x\right)+3\right]\)
3, = 4x(a-b)-6xy(a-b)
=2x(a-b)(2-3y)
4,
=3(2x+1)-(2x-5)(2x+1)
=(3-2x+5)(2x+1)
=(8-2x)(2x+1)
=2(4-x)(2x+1)
 

giúp mình câu 5,6.7.8. vs ạ

bạn viết câu hỏi dưới dạng trực quan để mn dễ hiểu nhé!

a) x³ - 64x = 0

x(x² - 64) = 0

x(x - 8)(x + 8) = 0

x = 0 hoặc x - 8 = 0 hoặc x + 8 = 0

*) x - 8 = 0

x = 8

*) x + 8 = 0

x = -8

Vậy x = -8; x = 0; x = 8

b) x³ - 4x² = -4x

x³ - 4x² + 4x = 0

x(x² - 4x + 4) = 0

x(x - 2)² = 0

x = 0 hoặc (x - 2)² = 0

*) (x - 2)² = 0

x - 2 = 0

x = 2

Vậy x = 0; x = 2

c) x² - 16 - (x - 4) = 0

(x - 4)(x + 4) - (x - 4) = 0

(x - 4)(x + 4 - 1) = 0

(x - 4)(x + 3) = 0

x - 4 = 0 hoặc x + 3 = 0

*) x - 4 = 0

x = 4

*) x + 3 = 0

x = -3

Vậy x = -3; x = 4

d) (2x + 1)² = (3 + x)²

(2x + 1)² - (3 + x)² = 0

(2x + 1 - 3 - x)(2x + 1 + 3 + x) = 0

(x - 2)(3x + 4) = 0

x - 2 = 0 hoặc 3x + 4 = 0

*) x - 2 = 0

x = 2

*) 3x + 4 = 0

3x = -4

x = -4/3

Vậy x = -4/3; x = 2

e) x³ - 6x² + 12x - 8 = 0

(x - 2)³ = 0

x - 2 = 0

x = 2

f) x³ - 7x - 6 = 0

x³ + 2x² - 2x² - 4x - 3x - 6 = 0

(x³ + 2x²) - (2x² + 4x) - (3x + 6) = 0

x²(x + 2) - 2x(x + 2) - 3(x + 2) = 0

(x + 2)(x² - 2x - 3) = 0

(x + 2)(x² + x - 3x - 3) = 0

(x + 2)[(x² + x) - (3x + 3)] = 0

(x + 2)[x(x + 1) - 3(x + 1)] = 0

(x + 2)(x + 1)(x - 3) = 0

x + 2 = 0 hoặc x + 1 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x + 1 = 0

x = -1

*) x - 3 = 0

x = 3

Vậy x = -1; x = -1; x = 3

Dòng cuối kết luận phải là \(\text{x }\in\text{ }\left\{-2;-1;3\right\}\) chứ ạ?

\(2x^2-x-6=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+\dfrac{3}{2}=0\end{matrix}\right.\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}3x-2=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

3x-2=0

3x=2

X=2/3

hoặc 4-x=0

X=4-0

X=4

\(\sqrt{x+2}+2\sqrt{4x+8}=15\)

\(\Leftrightarrow5\sqrt{x+2}=15\)

\(\Leftrightarrow x+2=9\)

hay x=7