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b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Ta có :
\(\left(2x+1\right)^3=27\)
\(\Rightarrow\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=3-1=2\)
\(\Rightarrow x=2:2=1\)
Vậy \(x=1\)
Ta có : \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
2x + 1 = 3
2x = 2
x = 1
Vậy x = 1
35-(x-7)=-7-(-27)+3x
=>35-x+7=-7+27+3x
=>-x+42=20+3x
=>3x-x=42-20
=>2x=22
=>x=11
x - ( 81 + 5x ) = 19
=> x - 81 - 5x = 19
=> ( x - 5x ) - 81 = 19
=> -4x = 19 + 81
=> -4x = 100
=> x = 100 : ( -4 )
=> x = -25
2x - ( 8 + 3x ) = 92
=> 2x - 8 - 3x = 92
=> ( 2x - 3x ) = 8 + 92
=> -x = 100
=> x = -100
1) Ta có: \(\left(-5+x\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5+x=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{5;7\right\}\)
2) Ta có: \(\left(30-x\right)\left(2x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}30-x=0\\2x-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-30\\2x=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=8\end{matrix}\right.\)
Vậy: \(x\in\left\{30;8\right\}\)
3) Ta có: \(\left(-5-x\right)\left(17+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-5-x=0\\17+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=5\\x=0-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-17\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;-17\right\}\)
4) Ta có: \(\left(-3x+18\right)\left(-5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+18=0\\-5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-18\\-5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-2\right\}\)
Bài nay ta có hai vế bạn hãy đặt giả sử một trong hai vế bằng 0 rồi giải phương trình cho mỗi vế bằng o
Bài 1:
a) Ta có: \(82-7\left(3x-4\right)=47\)
\(\Leftrightarrow82-21x+28-47=0\)
\(\Leftrightarrow-21x+63=0\)
\(\Leftrightarrow-21x=-63\)
hay x=3(nhận)
Vậy: x=3
b) Ta có: \(97+4\left(5x-7\right)=129\)
\(\Leftrightarrow97+20x-28-129=0\)
\(\Leftrightarrow20x-60=0\)
\(\Leftrightarrow20x=60\)
hay x=3(nhận)
Vậy: x=3
c) Ta có: \(\left(7x-13\right)\cdot27-12=15\)
\(\Leftrightarrow189x-351-12-15=0\)
\(\Leftrightarrow189x-378=0\)
\(\Leftrightarrow189x=378\)
hay x=2(nhận)
Vậy: x=2
d) Ta có: \(\left(2x+3\right)\cdot13+23=140\)
\(\Leftrightarrow26x+39+23-140=0\)
\(\Leftrightarrow26x-78=0\)
\(\Leftrightarrow26x=78\)
hay x=3(nhận)
Vậy: x=3
đ) Ta có: \(52x+8x-5x=70\)
\(\Leftrightarrow55x=70\)
\(\Leftrightarrow x=\frac{70}{55}\)(loại)
Vậy: x∈∅
e) Ta có: \(19x-3x-x=60\)
\(\Leftrightarrow15x=60\)
hay x=4(nhận)
Vậy: x=4
g) Ta có: \(7\left(3x+1\right)-5\left(3x+1\right)=74\)
\(\Leftrightarrow2\left(3x+1\right)=74\)
\(\Leftrightarrow3x+1=37\)
\(\Leftrightarrow3x=36\)
hay x=12(nhận)
Vậy: x=12
h) Ta có: \(5\left(3x-1\right)+7\left(3x-1\right)=96\)
\(\Leftrightarrow12\left(3x-1\right)=96\)
\(\Leftrightarrow3x-1=8\)
\(\Leftrightarrow3x=9\)
hay x=3(nhận)
Vậy: x=3
b: \(x\in\left\{4;-4\right\}\)
\(a,x^2=16\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ b,\left(2x-1\right)^3=-27\\ \Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\\ \Rightarrow2x-1=-3\\ \Rightarrow2x=-2\\ \Rightarrow x=-1\)