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a) \(\left(x-5\right)-\frac{1}{3}=\frac{2}{5}\)
\(\Rightarrow\left(x-5\right)=\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\left(x-5\right)=\frac{11}{15}\)
\(\Rightarrow x-5=\frac{11}{15}\)
\(\Rightarrow x=\frac{11}{15}+5\)
\(\Rightarrow x=\frac{86}{15}\)
b) \(\frac{2}{3}\cdot x-\frac{3}{2}\cdot x=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\)
\(\Rightarrow x\cdot\left(-\frac{5}{6}\right)=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}:\left(-\frac{5}{6}\right)\)
\(\Rightarrow x=-\frac{1}{2}\)
c) \(-\frac{2}{3}\cdot x+\frac{1}{5}=\frac{3}{10}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{3}{10}-\frac{1}{5}\)
\(\Rightarrow-\frac{2}{3}\cdot x=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}:\left(-\frac{2}{3}\right)\)
\(\Rightarrow x=-\frac{3}{20}\)
d) \(4-\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=-\frac{1}{5}\)
\(\Rightarrow\left(\frac{1}{2}\cdot x+\frac{3}{4}\right)=4-\left(-\frac{1}{5}\right)\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x+\frac{3}{4}=\frac{21}{5}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{21}{5}-\frac{3}{4}\)
\(\Rightarrow\)\(\frac{1}{2}\cdot x=\frac{69}{20}\)
\(\Rightarrow\)\(x=\frac{69}{20}:\frac{1}{2}\)
\(\Rightarrow\)\(x=\frac{69}{10}\)
(4x-12)(x3+64)=0
=> [x3+64=0=>x=4x-12=0=>4x=12=>x=3 olm bị lỗi nên em đừng có viết cách ra 1 quãng như kia nhé !
vậy x thuộc {3;4}
(3x-12)(x2-4)=0
=>[x2-4=0=>x2=4=>x=2 hoặc x=-23x-12=0=>3x=12=>x=4
vậy x thuộc {4;2;-2}
(x+3)3:3-1=-10
(x+3)3:3=-9
(x+3)3=-9.3
=>(x+3)3=-27
=>x+3=-3
=>x=-6
(3x-1)3-2=-66
(3x-1)3=-64
(3x-1)3=-43
=>3x-1=-4
=>3x=-3
=>x=-1
\(\left(4x-12\right)\left(x^3+64\right)=0\)
\(\Leftrightarrow4x-12=0\)
\(\Leftrightarrow4x=0+12\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=12\div4\)
\(\Leftrightarrow x=3\)
\(\Leftrightarrow x^3+64=0\)
\(\Leftrightarrow x^3=0=64\)
\(\Leftrightarrow x^3=\left(-64\right)\)
\(\Leftrightarrow x^3=\left(-4\right)^3\)
\(\Leftrightarrow x=\left(-4\right)\)
\(\Rightarrow x\in\left\{-4;3\right\}\)
\(\Leftrightarrow\left(3x-12\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow3x=0+12\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=12\div3\)
\(x=4\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=0+4\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2=\left(-2\right)^2\)
\(\Rightarrow x\in\left\{2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;2;4\right\}\)
Các câu khác tương tự nhé !
1. 3x - 36 = 12 . 75 + 25 . 12
3x - 36 = 12 . (75+25)
3x - 36 = 12 . 100
3x - 36 = 1200
3x = 1200 + 36
3x = 1236
=> x = 1236 : 3 = 412
câu 1 thôi nhá bạn
3x - 36 = 12 . 75 + 25 . 12
3x - 36 = 12 . (75 + 25)
3x - 36 = 1200
3x = 1164
x = 388
x : 2 - 12 = 33 . 40 + 33 . 59 + 33
x : 2 - 12 = 33 . 40 + 33 . 59 + 33 . 1
x : 2 - 12 = (40 + 59 + 1)
x : 2 - 12 = 3300
x : 2 = 3288
x = 1644
(x - 4) . (9 - x) = 0
Thỏa mãn điều kiện\(\hept{\begin{cases}x=4\\x=9\end{cases}}\)
(x - 6) . (2020 - x) = 0
Thỏa mãn điều kiện\(\hept{\begin{cases}x=6\\x=2020\end{cases}}\)
x . (6 - x) = 0
Thỏa mãn điều kiện 6 - x = 0
x = 6
(x - 3 - 12) . (20 - x) = 0
Thỏa mãn điều kiện \(x\le20\); 20 - x = 0
x = 20
a. (x2 - 4).(x+3/5) = 0
TH1: x2 - 4 = 0
x2 = 4
x2 = 22
-22
=> x = 2
-2
Vậy x \(\in\){-2;2}
\(a.\left[\left(6.x-39\right).7\right].4=12\)
\(\left(6.x-39\right).7=12:4\)
\(6.x-39=3:7\)
\(6.x=\frac{3}{7}+39\)
\(x=\frac{276}{7}:6\)
\(x=\frac{46}{7}\)
( câu a )
\(b.200-\left(2.x+6\right)=4^3\)
\(200-\left(2.x+6\right)=64\)
\(2.x+6=200-64\)
\(2.x=136-6\)
\(x=130:2\)
\(x=65\)
( câu b )
\(\left|x\right|=a\) <=> \(x=\pm a\)
\(\left|x+a\right|=7\)
<=> \(\orbr{\begin{cases}x+a=7\\x+a=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7-a\\x=-7-a\end{cases}}}\)
Vậy...
\(\left(-7\right)+\left|x-4\right|=-3\)
<=> \(\left|x-4\right|=4\)
<=> \(\orbr{\begin{cases}x-4=4\\x-4=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=0\end{cases}}}\)
Vậy...
\(13-\left|x+5\right|=10\)
<=> \(\left|x+5\right|=3\)
<=> \(\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}}\)
Vậy...
\(\left|x-10\right|-\left(-12\right)=4\)
<=> \(\left|x-10\right|=-8\) (vô lý)
=> vô nghiệm
\(-\left(12+4-10\right)-\left(4-x\right)=-\left(-3\right)\)
\(-6-\left(4-x\right)=3\)
\(\left(4-x\right)=-6-3\)
\(\left(4-x\right)=-9\)
\(x=4-\left(-9\right)\)
\(x=4+9\)
\(x=13\)