a) \(x^3+4x=0\)

\(\Rightarrow x\left(x^2+4\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2+4=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2=-4\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x\in\phi\end{array}\right.\)

Vậy: \(x=0\)

b) \(2\left(5-x\right)=4x-3\)

\(\Rightarrow10-2x=4x-3\)

\(\Rightarrow10+3=4x+2x\)

\(\Rightarrow13=6x\)

\(\Rightarrow x=\frac{13}{6}\)

x³+ 4x=0

<=> x(x²+4)=0

=> x=0 hoặc x²+4=0

Mà: x²+4 >4

=>x=0

b) (5-x).2=4x-3

<=>10-2x=4x-3

<=>10+3= 2x+4x

<=>13=6x

=>x=\(\frac{13}{6}\)

a) $(x-3)^2-(x+2)(x-2)=-5$

$\Rightarrow x^2-2\cdot x\cdot3+3^2-(x^2-2^2)=-5$

$\Rightarrow x^2-6x+9-(x^2-4)=-5$

$\Rightarrow x^2-6x+9-x^2+4=-5$

$\Rightarrow-6x+13=-5$

$\Rightarrow-6x=-18$

$\Rightarrow x=3$

b) $x^3-2x^2-4x+8=0$

$\Rightarrow(x^3-2x^2)-(4x-8)=0$

$\Rightarrow x^2(x-2)-4(x-2)=0$

$\Rightarrow (x^2-4)(x-2)=0$

$\Rightarrow (x^2-2^2)(x-2)=0$

$\Rightarrow (x-2)(x+2)(x-2)=0$

$\Rightarrow (x-2)^2(x+2)=0$

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

$\text{#}Toru$

Trả lời:

a, \(x^2-9-2\left(x-3\right)=0\)

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

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

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

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

Vậy x = 3; x = - 1 là nghiệm của pt.

b, \(x\left(x-5\right)-4x+20=0\)

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

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

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

Vậy x = 5; x = 4 là nghiệm của pt.

c, \(2x^2+3x-5=0\)

\(\Leftrightarrow2x^2+5x-2x-5=0\)

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

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

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

Vậy x = - 5/2; x = 1 là nghiệm của pt.

CHỊU KHÓ

TL

a) pt tương đương:
x2−81−x2+6x−9

=0⇔6x

=90⇔x=15

b)

x=4,

x=5

c)

x=-5/2,

x=1

HT

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

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

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

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

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

Vậy .....

a, 25x^2 - 1 - (5x -1)(x+2)=0

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

=> (5x-1)(5x+1) + (5x-1)(x+2) = 0

=> (5x-1)(5x+1+x+2) = 0

=> (5x-1)(6x+3) = 0

=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)

a)     25x² - 1 - ( 5x - 1 )( x + 2 ) = 0

<=> ( 5x - 1 )( 5x + 1 ) - ( 5x - 1 )( x + 2 ) = 0

<=> ( 5x - 1 )( 5x + 1 - x - 2) = 0

<=> ( 5x - 1 )( 4x - 1 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{4}\end{cases}}}\)

Vậy .......

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

a) \(\Rightarrow x\left(x+3\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

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

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

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

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy giỏi zữ vậy

a) (x - 4)² - 36 = 0

=> (x - 4)² = 36

=> x - 4 = 6 hoặc x - 4 = -6

=> x = 10 hoặc x = -2

b) hình như sai đề bn ạ

c) x(x - 5) - 4x + 20 = 0

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

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

=> x - 5 = 0 hoặc x - 4 = 0

=> x = 5 hoặc x = 4

a) x² - 4x - 5 = 0

=> x² - 5x + x - 5 = 0

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

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

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

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

b) 4x² + 7x - 11 = 0

=> 4x² + 11x - 4x - 11 = 0

=> x(4x + 11) - (4x + 11) = 0

=> (x - 1)(4x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)

c) -7x² + 6x + 1 = 0

=> -7x² + 7x - x + 1 = 0

=> -7x(x - 1) - (x - 1) = 0

=> (-7x - 1)(x - 1) = 0

=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)

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

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

d) -10x² + 7x + 3 = 0

=> -10x² + 10x - 3x + 3 = 0

=> -10x(x - 1) - 3(x - 1) = 0

=> (-10x - 3)(x - 1) = 0

=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

\(a,x^2-4x-5=0\)

\(\Rightarrow x^2-5x+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

\(b,4x^2+7x-11=0\)

\(\Rightarrow4x^2-4x+11x-11=0\)

\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)

\(c,-7x^2+6x+1=0\)

\(\Rightarrow-7x^2+7x-x+1=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

\(d,-10x^2+7x+3=0\)

\(\Rightarrow-10x^2+10x-3x+3=0\)

\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)

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