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a, 2.(4x-3)-3(x+5)+4(x-10)=5(x+2)
2.4x-2.3-3.x+3.5+4x-4.10=5x+5.2
8x-6-3x+15+4x-40=5x-10
8x-3x+4x-5x-6-15-40-10=0
4x-71=0
4x=71
x=71:4
x=71/4
x + 5/2 . x - 3/2 = 9/4
<=> x( 1+ 5/2 ) - 3/2 = 9/4
<=> x . 7/2 = 9/4 + 3/2
<=> x .7/2 = 15/4
<=> x = 15/4 : 7/2
<=> x = 15/14
TA CÓ:
X + 5/2 . X - 3/2 = 9/4
X + 5/2 .X = 9/4 +3/2 = 15/4
(X . 1) + (5/2 . X) = 15/4
X . (1 + 5/2) =15/4
X . 7/2 = 15/4
X = (15/4) / (7/2)
X = 15/14
DỄ ÒM MÀ
BẠN HỌC TRỪNG NÀO MÀ MAI NỘP VẬY
\(\dfrac{-5}{3}-\left(\dfrac{5}{12}-\dfrac{3}{4}\right)< x< \dfrac{11}{6}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{-5}{3}-\left(\dfrac{5}{12}-\dfrac{3}{4}\right)\\x< \dfrac{11}{6}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{-5}{3}-\dfrac{5}{12}+\dfrac{3}{4}\\x< \dfrac{11}{6}-\dfrac{1}{3}-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{-20}{12}-\dfrac{5}{12}+\dfrac{9}{12}\\x< \dfrac{22}{12}-\dfrac{4}{12}-\dfrac{3}{12}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>-\dfrac{4}{3}\\x< \dfrac{5}{4}\end{matrix}\right.\Rightarrow x\in\left\{-\dfrac{4}{3};\dfrac{5}{4}\right\}}\)
\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=2\)
\(\le\left|x\right|-\left|2\right|+\left|x\right|-\left|3\right|+\left|x\right|-\left|4\right|\)
\(x-2+x-3+x-4=2\)
\(\Leftrightarrow x+x+x-2-3-4=2\)
\(\Leftrightarrow x^3-2-3-4=2\)
\(x^3=2+4+3+2\)
\(x^3=11\)
\(x=\sqrt[3]{11}\)
a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)
b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)
c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)
\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)
Vì -50<-30<4<5
nên A<D<B<C
a: \(A=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}=\dfrac{-25}{60}=\dfrac{-50}{120}\)
b: \(B=\dfrac{3}{4}\cdot\dfrac{1}{12}\cdot\dfrac{2}{3}=\dfrac{1}{24}=\dfrac{5}{120}\)
c: \(C=\dfrac{5}{4}\cdot\dfrac{1}{15}\cdot\dfrac{2}{5}=\dfrac{2}{60}=\dfrac{1}{30}=\dfrac{4}{120}\)
\(D=-3\cdot\dfrac{-7}{12}\cdot\dfrac{1}{-7}=-\dfrac{1}{4}=\dfrac{-30}{120}\)
Vì -50<-30<4<5
nên A<D<B<C
a) \(\frac{3}{4}-\frac{2}{5}.x=x\)
\(\Rightarrow\frac{-2}{5}.x-x=\frac{-3}{4}\)
\(x.\left(\frac{-2}{5}-1\right)=\frac{-3}{4}\)
\(x.\frac{-7}{5}=\frac{-3}{4}\)
\(x=\frac{-3}{4}:\left(\frac{-7}{5}\right)\)
\(x=\frac{15}{28}\)
b) (2x-1).(3x-1/5).(4-2x) = 0
=> 2x - 1 = 0 => 2x = 1 => x = 1/2
3x-1/5 = 0 => 3x = 1/5 => x = 1/15
4-2x = 0 => 2x = 4 => x = 2
KL: x = 1/2 hoặc x = 1/15 hoặc x = 2
a)\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\frac{2}{5}+x=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{2}{5}\)
\(x=-\frac{3}{20}\)
Vậy \(x=-\frac{3}{20}\)
b)\(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{3}\)
\(\frac{1}{4}:x=-\frac{1}{12}\)
\(x=\frac{1}{4}:\left(-\frac{1}{12}\right)\)
\(x=-3\)
Vậy \(x=-3\)
yeah thank you !