Tìm x biết : (2x-7)3=8(7-2x)2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( x - 1 )( x + 2 ) - x - 2 = 0
<=> ( x - 1 )( x + 2 ) - ( x + 2 ) = 0
<=> ( x + 2 )( x - 2 ) = 0
<=> x = ±2
( 2x - 7 )3 = 8( 7 - 2x )2
<=> ( 2x - 7 )3 - 8( 2x - 7 )2 = 0
<=> ( 2x - 7 )2( 2x - 15 ) = 0
<=> x = 7/2 hoặc x = 15/2
1: 7-x=8+(-7)
=>7-x=8-7=1
=>x=7-1=6
2: \(x-8=\left(-3\right)-8\)
=>x-8=-11
=>\(x=-11+8=-3\)
3: \(2-x=10-9+23\)
=>\(2-x=33-9=24\)
=>x=2-24=-22
4: \(-2-x=15\)
=>\(x=-2-15=-17\)
5: \(-7+x-8=-3-1+13\)
=>x-14=13-4=9
=>x=9+14=23
6: 100-x+7=-x+3
=>107-x=3-x
=>107=3(vô lý)
7: \(23+x=8-2x\)
=>\(x+2x=8-23\)
=>3x=-15
=>x=-15/3=-5
\(\Leftrightarrow\left(2x-3\right)^7\left(2x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
a) \(\left(x+1\right)^2=3\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)^2-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
b) \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)
\(\Leftrightarrow\left(2x-7\right)^3-8\left(2x-7\right)^2=0\)
\(\Leftrightarrow\left(2x-7\right)^2\left(2x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-7\right)^2=0\\2x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=15\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{15}{2}\end{cases}}\)
a, \(\left(x+1\right)^2=3\left(x+1\right)\Leftrightarrow x^2+2x+1=3x+3\)
\(\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
b, \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)
\(\Leftrightarrow8x^3-116x^2+518x-735=0\Leftrightarrow\orbr{\begin{cases}x=3,5\\x=7,5\end{cases}}\)
a. \(\dfrac{8}{7}-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=-1\)
\(-\dfrac{1}{7}:\left(\dfrac{x}{3}-2\right)=\dfrac{15}{7}\)
\(\dfrac{x}{3}-2=\dfrac{-1}{15}\)
\(\dfrac{x}{3}=\dfrac{29}{15}\)
\(x=5,8\)
b. \(\dfrac{5}{8}+\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{4}\)
\(\dfrac{1}{4}\left(2x-1\right)=\dfrac{5}{8}\)
\(2x-1=\dfrac{5}{2}\)
\(2x=\dfrac{7}{2}\)
\(x=\dfrac{7}{4}\)
a) 5(2x -1) - 4(8 - 3x) = 7
<=> 10x - 5 - 32 + 12x = 7
<=> 22x = 44
<=> x =2
Vậy x = 2 là nghiệm phương trình
b) 7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x - 2) - (x + 4)
<=> 14x - 35 - 35x + 10 + 10x - 14 = x - 2 - x - 4
<=> -11x - 39 = - 6
<=> -11x = 33
<=> x = -3
Vậy x = -3 là nghiệm phương trình
\(a,10x-5-32+12x=7\)
\(22x=44\)
\(x=2\)
\(b,14x-35-35x+10+10x-14=x-2-x-4\)
\(-11x-39=-6\)
\(-11x=-33\)
\(x=3\)
\((2x-7)^3=8(7-2x)^2\)
⇔ \((2x-7)^3=8(2x-7)^2\) (*)
\(TH1: (2x-7)^2=0\)
Khi đó: \(2x-7=0\) ⇔ \(x=\dfrac{7}{2} \)
\(TH2:\left(2x-7\right)^2\ne0\)
Khi đó: (*) ⇔ \(2x-7=8\) (chia 2 vế cho \((2x-7)^2\))
⇔ \(x=\dfrac{15}{2} \)
Vậy \(x=\dfrac{15}{2}\); \(x=\dfrac{7}{2}\)
chuyển vế sang là thành \(-8(2x-7)^2 \) chứ ạ