Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

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

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

= x(x - 1 - y)(x - 1 + y)

= x(x - y - 1)(x + y - 1)

e) 27 - 3x²

= 3(9 - x²)

= 3(3 - x)(3 + x)

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

= 2(x + 1 - y)(x + 1 + y)

= 2(x - y + 1)(x + y + 1)

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

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

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

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

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

\(=5xy\left(x-4y\right)\)

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

\(=\left(4x-1\right)^2-y^2\)

\(=\left(4x-1-y\right)\left(4x-1+y\right)\)

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

\(=x\left(x^2-2x+1-y^2\right)\)

\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)

\(=x\left[\left(x-1\right)^2-y^2\right]\)

\(=x\left(x-1-y\right)\left(x-1+y\right)\)

e: \(27-3x^2\)

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

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

f: \(2x^2+4x+2-2y^2\)

\(=2\left(x^2+2x+1-y^2\right)\)

\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)

\(=2\left[\left(x+1\right)^2-y^2\right]\)

\(=2\left(x+1+y\right)\left(x+1-y\right)\)

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) x² + 2x - 3x - 6 = 0

(x² + 2x) - (3x + 6) = 0

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

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

x + 2 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x - 3 = 0

x = 3

Vậy x = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

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

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

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

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

Bài 1:

\(A=-x^2-2x+9\)

\(A=-\left(x^2+2x-9\right)\)

\(A=-\left(x^2+2x+1-10\right)\)

\(A=-\left(x+1\right)^2+10\)

Vì \(-\left(x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x+1\right)^2+10\le10\)

\(\Rightarrow Amax=10\Leftrightarrow x=-1\)

\(B=-9x^2+6x+25\)

\(B=-\left(9x^2-6x-25\right)\)

\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)

\(B=-\left(3x-1\right)^2+26\)

Vì \(-\left(3x-1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(3x-1\right)^2+26\le26\)

\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(C=-x^2+x+1\)

\(C=-\left(x^2-x-1\right)\)

\(C=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1\right)\)

\(C=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)

Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(D=-2x^2+3x+1\)

\(D=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)

\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)

\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)

Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x

\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)

\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)

\(E=-25x^2-10x+7\)

\(E=-\left(25x^2+10x-7\right)\)

\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)

\(E=-\left(5x+1\right)^2+8\)

Vì \(-\left(5x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(5x+1\right)^2+8\le8\)

\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

Bài 2:

\(A=9x^2+6x+4\)

\(A=\left(3x\right)^2+2.3x+1+3\)

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

Vì \(\left(3x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(3x+1\right)^2+3\ge3\)

\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)

\(B=4x^2+4x+12\)

\(B=\left(2x\right)^2+2.2x+1+11\)

\(B=\left(2x+1\right)^2+11\)

Vì \(\left(2x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x+1\right)^2+11\ge11\)

\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)

\(C=x^2+x+3\)

\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

\(D=2x^2+3x+1\)

\(D=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)

\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)

\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)

Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)

\(E=64x^2+16x+3\)

\(E=\left(8x\right)^2+2.8x+1+2\)

\(E=\left(8x+1\right)^2+2\)

Vì \(\left(8x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(8x+1\right)^2+2\ge2\)

\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

\(n,3x^3-3x^2-6x=0\)

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

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

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)

\(m,x^2-11x+28=0\)

\(\Leftrightarrow x^2-4x-7x+28=0\)

\(\Leftrightarrow x\left(x-4\right)-7\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=7\end{cases}}\)

\(l,\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)

\(\Leftrightarrow16x^2+24x+9=4x^2-8x+4\)

\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)

\(\Leftrightarrow12x^2+32x+5=0\)

\(\Leftrightarrow\left(x+\frac{1}{6}\right)\left(x+\frac{5}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{2}\end{cases}}\)

1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)

\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)

\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)

\(\Leftrightarrow5x-6=0\)

\(\Leftrightarrow5x=6\)

\(\Leftrightarrow x=\frac{6}{5}\)

Vậy: \(x=\frac{6}{5}\)

2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)

\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)

\(\Leftrightarrow12x+24=0\)

\(\Leftrightarrow12x=-24\)

\(\Leftrightarrow x=-2\)

Vậy: x=-2

3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)

\(\Leftrightarrow15x-30=0\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=2\)

Vậy: x=2

4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)

\(\Leftrightarrow83x-83=0\)

\(\Leftrightarrow83x=83\)

\(\Leftrightarrow x=1\)

Vậy: x=1

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4