Question

Tìm x, biết:

a) x² + 10x = 24

b) 4x² + 4x = 24

c) 4x² - 4x = 48

Các bạn giúp mình với nhé!!! Mình đang rất cần

Answer 1

a)\(x^2+10x=24\)

\(\Leftrightarrow x^2+10x-24=0\)

\(\Leftrightarrow x^2-2x+12x-24=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)

b)\(4x^2+4x=24\)

\(\Leftrightarrow4x^2+4x-24=0\)

\(\Leftrightarrow4\left(x^2+x-6\right)=0\)

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

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

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

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

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

c)\(4x^2-4x=48\)

\(\Leftrightarrow4x^2-4x-48=0\)

\(\Leftrightarrow4\left(x^2-x-12\right)=0\)

\(\Leftrightarrow x^2-x-12=0\)

\(\Leftrightarrow x^2+3x-4x-12=0\)

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

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

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

Answer 2

\(a,x^2+10x=24\)
\(\Leftrightarrow x^2+10x-24=0\)
\(\Leftrightarrow x^2-2x+12x-24=0\)
\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)
\(\text{Vậy x=2 hoặc x=-12 }\)
\(b,4x^2+4x=24\)
\(\Leftrightarrow4x^2+4x-24=0\)
\(\Leftrightarrow4x^2-8x+12x-24=0\)
\(=4x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+12=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
Vậy hoặc \(\text{Vậy x=2 hoặc x=-3 }\)
\(c,4x^2-4x=48\)
\(\Leftrightarrow4x^2-4x-48=0\)
\(\Leftrightarrow\left[\left(2x\right)^2-2.2x+1^2\right]-1^2-48=0\)
\(\Leftrightarrow\left(2x-1\right)^2-49=0\)
\(\Leftrightarrow\left(2x-1\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-1-7\right)\left(2x-1+7\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-8=0\\2x+6=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(\text{Vậy x=4 hoặc x=-3 }\)

Answer 3

bạn chuyển vế tất cả bên phải sang bên trái rồi giải theo vi-et hoặc đưa về pt tích là OK..:)

 

Answer 4

a. x²+10x=24

<=>x²+10x-24=0

<=>(x²+12x)-(2x+24)=0

<=>x(x+12)-2(x+12)=0

<=>(x-2)(x+12)=0

=>x-2=0<=>x=2; x+12=0<=>x=-12

Tương tự:b.x=2;-3

c.x=4;-3

Answer 5

a)

x²+10x=24

=> x²+10x-24=0

=> x²+10x+25-49=0

=> (x+5)²-49=0

=> (x+5+7)(x+5-7)=0

=> (x+12)(x-2)=0

=>\(\left[\begin{array}{nghiempt}x+12=0\\x-2=0\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=-12\\x=2\end{array}\right.}\)

b)

4x²+4x=24

=> 4x²+4x+1-25=0

=> (2x+1)²-25=0

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

=> (2x+6)(2x-4)=0

=>\(\left[\begin{array}{nghiempt}2x+6=0\\2x-4=0\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x+3=0\\x-2=0\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}x=-3\\x=2\end{array}\right.}\)

c)4x²-4x=48

=> 4x²-4x+1-49=0

=> (2x-1)²-49=0

=> (2x-1-7)(2x-1+7)=0

=> (2x-8)(2x+6)=0

=>\(\left[\begin{array}{nghiempt}2x+6=0\\2x-8=0\end{cases}\Rightarrow}\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.}\)