a) \(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9\)

\(\Rightarrow\left(x+2\right)^2=3^2\)

\(\Rightarrow x+2=3\)

\(\Rightarrow x=3-2=1\)

a) ( x + 2 )² = 9

=> ( x + 2 ) ² = 9

=> ( x + 2 )² = 3²

=> x + 2 = + 3

=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

Vậy x = -1; 5

b) ( x + 2 )² - x² + 4 = 0

=> ( x + 2 )² - ( x² - 4 ) = 0

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

=> ( x + 2 ) ( x + 2 -  x + 2 ) = 0

=> ( x + 2 ) . 4 = 0

=> x + 2 = 0 

=> x = - 2

Vậy x = - 2 

c)  5 ( 2x - 3 )² - 5 ( x + 1 )² - 15( x + 4 ) ( x - 4 )  = - 10

=> 5 ( 4x² - 12x + 9 ) - 5 ( x² + 2x + 1 ) - 15 ( x² - 4² ) = - 10

=> 20x² - 60x + 45 - 5x² - 10x - 5 - 15x² + 240 = -10

=> - 70x + 280 = - 10

=> - 70x = - 290

=> x = \(\frac{29}{7}\)

Vậy x = \(\frac{29}{7}\)

d)  x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x² - 2x + 4 ) = 3

=> x ( x² - 25 ) - ( x³ - 8 ) = 3

=> x³ - 25x - x³ + 8 = 3

=> - 25x + 8 = 3

=> - 25x = -5

=> x = \(\frac{1}{5}\)

Vậy x = \(\frac{1}{5}\)

\(a,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[x^2-2x-3-x^2+3x-10\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases};x-13=0}\)

\(\Leftrightarrow x=1;x=2\)hoặc \(x=13\)

\(b,\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)+6\left(x^2+x\right)-12=0\)

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

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

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

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

Lại do \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+5\frac{3}{4}\ge5\frac{3}{4}>0\)

\(\Rightarrow\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)=0\)

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

c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)

\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)

Nên x + 2009 = 0 => x = -2009

a) x³ - 7x + 6 = 0

x³ - x - 6x + 6 = 0

(x³ - x) - (6x - 6) = 0

x(x² - 1) - 6(x - 1) = 0

x(x - 1)(x + 1) - 6(x - 1) = 0

(x - 1)[x(x + 1) - 6] = 0

(x - 1)(x² + x - 6) = 0

(x - 1)(x² - 2x + 3x - 6) = 0

(x - 1)[(x² - 2x) + (3x - 6)] = 0

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

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

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

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x + 3 = 0

x = -3

Vậy x = -3; x = 1; x = 2

a: \(x^3-7x+6=0\)

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

=>\(x\left(x^2-1\right)-6\left(x-1\right)=0\)

=>x(x-1)(x+1)-6(x-1)=0

=>(x-1)(x^2+x-6)=0

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

=>\(\left[\begin{array}{l}x-1=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\\ x=2\end{array}\right.\)

b: \(x^4+4x^2-5=0\)

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

=>\(\left(x^2+5\right)\left(x^2-1\right)=0\)

=>\(x^2-1=0\)

=>\(x^2=1\)

=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

c: \(x^4+x^3-x^2-x=0\)

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

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

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

=>\(\left[\begin{array}{l}x=0\\ x+1=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-1\\ x=1\end{array}\right.\)

d: \(x^2+6x-x-6=0\)

=>x(x+6)-(x+6)=0

=>(x+6)(x-1)=0

=>\(\left[\begin{array}{l}x+6=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=1\end{array}\right.\)

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

=>x(x-4)+5(x-4)=0

=>(x-4)(x+5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-5\end{array}\right.\)

f: \(x^2-10x+2x-20=0\)

=>x(x-10)+2(x-10)=0

=>(x-10)(x+2)=0

=>\(\left[\begin{array}{l}x-10=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10\\ x=-2\end{array}\right.\)

g: \(x^4-x^3-x^2+1=0\)

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

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

=>\(\left(x-1\right)\left(x^3-x-1\right)=0\)

TH1: x-1=0

=>x=1

TH2: \(x^3-x-1=0\)

=>x≃1,32

h: \(x^5+x^4+x^3+x^2+x+1=0\)

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

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

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

nên \(x^3+1=0\)

=>\(x^3=-1\)

=>x=-1

i: \(x^2-9+\left(x+3\right)\left(3x-5\right)=0\)

=>(x-3)(x+3)+(x+3)(3x-5)=0

=>(x+3)(x-3+3x-5)=0

=>(x+3)(4x-8)=0

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

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

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

j: \(64x^2-9+8x+3=0\)

=>(8x+3)(8x-3)+(8x+3)=0

=>(8x+3)(8x-3+1)=0

=>(8x+3)(8x-2)=0

=>\(\left[\begin{array}{l}8x+3=0\\ 8x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac38\\ x=\frac28=\frac14\end{array}\right.\)

giải

a)4x^2-20x-(4x^2+3x-4x-3)=5

4x^2-20x-4x^2-3x+4x+3=5

-19x+3=5

-19x=5-3

-189x=2

x=-2/19

mik giải luôn đó chứ ko viết đầu bài đâu

c)

2x(x-3)-2(x^2-4)=4

2x^2-6x-2x^2+8=4

-6x+8=44

-6x=4-8

-6x=-4

x=2/3

a) x³ + 3x² + 3x + 1 = 64

=> (x + 1)³ = 64

=> (x + 1)³ = 4³

=> x + 1 = 4 => x = 3

b) x³ + 6x² + 9x = 4x

=> x³ + 6x² + 9x - 4x = 0

=> x³ + 6x² + 5x = 0

=> x³ + 5x² + x² + 5x = 0

=> x²(x + 5) + x(x + 5) = 0

=> (x + 5)(x² + x) = 0

=> (x + 5)x(x + 1) = 0

=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)

c) 4(x - 2)² = (x + 2)²

=> 4(x² - 4x + 4) = x² + 4x + 4

=> 4x² - 16x + 16 = x² + 4x + 4

=> 4x² - 16x + 16 - x² - 4x - 4 = 0

=> 3x² - 20x + 12 = 0

=> 3x² - 18x - 2x + 12 = 0

=> 3x(x - 6) - 2(x - 6) = 0

=> (x - 6)(3x - 2) = 0

=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)

d) x⁴ - 16x² = 0

=> x²(x² - 16) = 0

=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)

e) x⁴ - 4x³ + x² - 4x = 0

=> x⁴ + x² - 4x³ - 4x = 0

=> x²(x² + 1) - 4x(x² + 1) = 0

=> (x² - 4x)(x² + 1) = 0

=> x(x - 4)(x² + 1) = 0

=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x² + 1 \(\ge\)1 > 0 \(\forall\)x)

f) x³ + x = 0 => x(x²  + 1) = 0 => x = 0 (vì x² + 1 \(\ge1>0\forall\)x)

\(a,x^3+3x^2+3x+1=64\)

\(\left(x+1\right)^3=64\)

\(\left(x+1\right)^3=4^3\)

\(x+1=4\)

\(x=3\)

làm nốt

d) (2x-1)(3x+2)(3-x)

=(6x²+x-2)(3-x)

=-6x³+17x²+5x-6

e) (x+3)(x²+3x-5)

=x³+6x²+4x-15

f) (xy-2)(x³-2x-6)

=x⁴y-2x³-2x²y-6xy+4x+12

g) (5x³-x2+2x-3)(4x²-x+2)

=20x⁵-9x⁴+19x³-16x²+7x-6

Bài 1:

a) (x-2)(x2+3x+4)

=x(5x+4)-2(5x+4)

= 5x²+4x-10x-8

=5x²-6x-8