\(\left(x+4\right)^2-4x=\left(x-3\right)\left(x+3\right)-11\)

\(\Leftrightarrow x^2+8x+16-4x=x^2-9-11\)

\(\Leftrightarrow\left(x^2-x^2\right)+4x+\left(16+9+11\right)=0\)

\(\Leftrightarrow4x=-36\)

\(\Leftrightarrow x=-9\)

Vậy \(x=-9\)

\(\Leftrightarrow x^2+8x+16-4x=x^2-9-11\)

\(\Leftrightarrow x^2-x^2+8x-4x=-9-11-16\)

\(\Leftrightarrow4x=-36\)

\(\Leftrightarrow x=-9\)

a) Ta có : 

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

\(\Leftrightarrow3x=3x+2\)

\(\Leftrightarrow0=2\) ( vô lí )

Do đó pt đã cho vô nghiệm

b) Ta có  \(\left|x\right|=-x^2-2\) (1)

Nhân xét : VT (1) : \(\left|x\right|\ge0\forall x\)

VP (1) : \(-x^2\le0\Leftrightarrow-x^2-2\le-2\forall x\)

Do đó : \(VT\ne VP\)

Vì vậy pt đã cho vô nghiệm

(x-1)²-9=0

( x-1)²= 0+9 

(x-1)²=9

( x-1)²= 3²

=> x-1= 3

x= 3+1

x= 4

Vậy.....

bài này k đến lớp 8 đâu

\(\left(\text{x-1}\right)^2-9=0\)

<=>\(\left(x-1\right)^2=9\)

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

<=>x=4

(x-10)²-125=x(x-15)-5

(x−10)^2−125=x(x−15)−5

Step 1: Simplify both sides of the equation.

x^2−20x−25=x^2−15x−5

<=>x^2−20x−25−x^2=x^2−15x−5−x^2

−20x−25=−15x−5

<=>−20x−25+15x=−15x−5+15x

−5x−25=−5

<=>−5x−25+25=−5+25

−5x=20
\(\frac{-5x}{-5}=\frac{20}{-5}\)

=> x=-4

\(P=\frac{x+1}{2x}\Rightarrow Q=2P=\frac{2x+2}{2x}=1+\frac{2}{2x}\)

\(\Rightarrow Q=2P\in Z\Leftrightarrow1+\frac{2}{2x}\in Z\Leftrightarrow\frac{2}{2x}\in Z\)

\(\Leftrightarrow2x\inƯ\left(2\right)=\left\{-1;1;-2;2\right\}\)

\(\Leftrightarrow x\in\left\{-\frac{1}{2};\frac{1}{2};-1;1\right\}\)

Mà \(x\in Z\Rightarrow x\in\left\{-1;1\right\}\)

...

\(6x^2-7x^2-16x+m=0\)

\(\Leftrightarrow-x^2-16x+m=0\)

Nếu pt có 1 nghiệm bằng 1 thì \(-1-16+m=0\Rightarrow m=17\)

Phương trình trở thành \(-x^2-16x+17=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+17\right)=0\)

Suy ra nghiệm còn lại của pt là -17

-4kx + 2 = k - 1

x = -3 

<=> 12k + 2 = k - 1

<=> 11k = - 3

<=> k = -3/11

Vs \(x=-3\)

\(-4kx+2=k-1\Rightarrow-4k.\left(-3\right)+2=k-1\)

\(12k+2=k-1\)

\(12k-k=-1-2\)

\(11k=-3\Leftrightarrow k=-\frac{3}{11}\)

a, \(5\left(m+3x\right)\left(x+1\right)-4\left(1+2x\right)=80\)

Phương trình nhận \(x=2\)làm nghiệm nên :

\(5\left(m+3.2\right)\left(2+1\right)-4\left(1+2.2\right)=80\)

\(\Leftrightarrow15m+90-20=80\)

\(\Leftrightarrow15m=80+20-90\)

\(\Leftrightarrow15m=10\Leftrightarrow m=1,5\)

....

b, \(3\left(2x+m\right)\left(3x+2\right)-2\left(3x+1\right)^2=43\)

Phương trình nhận \(x=1\)làm nghiệm nên :

\(3\left(2.1+m\right)\left(3.1+2\right)-2\left(3.1+1\right)^2=43\)

\(\Leftrightarrow30+15m-32=43\)

\(\Leftrightarrow15m=43+32-30\)

\(\Leftrightarrow15m=45\Leftrightarrow m=3\)

....

\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)

\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)

\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)

\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow416-x=0\)

\(\Leftrightarrow x=416\)

a) 5(m + 3x)(x + 1) - 4(1 + 2x) = 80

Phương trình có nghiệm x = 2:

5(m + 3.2)(2 + 1) - 4(1 + 2.2) = 80

<=> 5(m + 6).3 - 4.5 = 80

<=> 15(m + 6) - 4.5 = 80

<=> 15(m + 6) - 20 = 80

<=> 15(m + 6) = 80 + 20

<=> 15(m + 6) = 100

<=> m + 6 = 100 : 15

<=> m + 6 = 20/3

<=> m = 20/3 - 6

<=> m = 2/3

b) 3(2x + m)(3x + 2) - 2(3x + 1)² = 43

Phương trình có nghiệm x = 1:

3(2.1 + m)(3.1 + 2) - 2(3.1 + 1)² = 43

<=> 3(2 + m).5 - 2.16 = 43

<=> 15(2 + m) - 32 = 43

<=> 15(2 + m) = 43 + 32

<=> 15(2 + m) = 75

<=> 2 + m = 75 : 15

<=> 2 + m = 5

<=> m = 5 - 2

<=> m = 3

a, \(5x+5=5x+5\)

\(0x=0\)

\(\RightarrowĐPCM\)

b, \(x^2+8x+16=x^2+8x+16\)

\(0x=0\)

\(\RightarrowĐPCM\)

a, \(5\left(x+1\right)=5x+5\)

\(\Leftrightarrow5x+5=5x+5\)

Vậy phương trình đúng với mọi nghiệm \(x\in R\)

b,\(\left(x+4\right)^2=x^2+8x+16\)

\(\Leftrightarrow x^2+8x+16=x^2+8x+16\)

Vậy phương trình đúng với mọi nghiệm \(x\in R\)

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

<=> x³ - 6x² + 12x - 8 + 9x² - 1 - x³ - 3x² - 3x - 1 = 0

<=> 9x - 10 = 0

<=> 9x = 10

<=> x = 10/9

Vậy S = {10/9}

b) (x + 1)(2x - 3) = (2x - 1)(x + 5)

<=> 2x² - x - 3 - 2x² - 9x + 5 = 0

<=> -10x + 2 = 0

<=> -10x = -2

<=> x = 1/5

Vậy S = {1/5}

c) (x - 1)³ - x(x + 1)² = 5x(2 - x) - 11(x + 2)

<=> x³ - 3x² + 3x - 1 - x³ - 2x² - x = 10x - 5x² - 11x - 22

<=> -5x² + 2x + 5x² + x + 22 - 1 = 0

<=> 3x = -21

<=> x = -7

Vậy S = {-7}

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

<=> x² + x - 12 - 6x + 4 - x² + 8x - 16 = 0

<=> 3x - 24 = 0

<=> 3x = 24

<=> x = 8

Vậy S = {8}

e) x(x + 3)² - 3x = (x + 2)³ + 1

<=> x³ + 6x² + 9x - 3x = x³ + 6x² + 12x + 8 + 1

<=> x³ + 6x² + 6x - x³ - 6x² - 12x = 9

<=> -6x = 9

<=> x = -3/2

Vậy S = {-3/2}

f) (x + 1)(x² - x + 1) - 2x = x(x + 1)(x- 1)

<=> x³ + 1 - 2x = x³ - x

<=> x³ - 2x - x³ + x = -1

<=> -x = -1

<=> x = 1

Vậy S = {1}