\(3x+5-2x-8=4x+\frac{1}{2}\)
\(3x+5-2x-8-4x-\frac{1}{2}=0\)
\(\left(3x-2x-4x\right)+\left(5-8-\frac{1}{2}\right)=0\)
\(-3x-\frac{7}{2}=0\)
\(-3x=\frac{7}{2}\)
\(x=\frac{-7}{6}\)
\(3x+5-2x-8=4x+\frac{1}{2}\)
\(3x+5-2x-8-4x-\frac{1}{2}=0\)
\(\left(3x-2x-4x\right)+\left(5-8-\frac{1}{2}\right)=0\)
\(-3x-\frac{7}{2}=0\)
\(-3x=\frac{7}{2}\)
\(x=\frac{-7}{6}\)
tìm x
\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{2}x-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=\frac{22}{45}x^2\)
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1.Tìm số hữu tỉ x:
a)\(4^{x+2}.3^x=16.12^5\)\(\)
b)\(\frac{3}{2}.4x+\frac{5}{3}.4^{x+2}==\frac{3}{2}.4^8+\frac{5}{3}.4^{10}\)
Tìm x, biết :
\(a.\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(b.\left(4x-3\right)^4=\left(4x-3\right)^2\)
tìm x
a) \(|\frac{3}{2}x+\frac{1}{2}|=|4x-1|\) b) \(|\frac{5}{4}x-\frac{7}{2}|-|\frac{5}{8}x-\frac{7}{2}|=0\)
tìm x
\(\frac{x+2}{4x-1}=\frac{x-5}{4x+1}\)
Câu 1: tìm x,y,z biết rằng:
a) \(\frac{1}{3}-|\frac{5}{4}-2x|=\frac{1}{4}\)
b)\(|5x-4=|x+2|\)
c) \(||2x-1|+\frac{1}{2}|=\frac{4}{5}\)
d) \(|x+1|+|x+2|+|x+3|=4x\)
Câu 2 Cho A=\(\frac{\sqrt{x+1}}{\sqrt{x-3}}\)tìm số nguyên x để A có giá trị là số nguyên
tìm x biết
a,\( |2-\frac{3}{2}x|-4=x+2\)
b,\(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
c,\(2.3^x-405=3^{x-1}\)
d.\(\left(\frac{1}{81}\right)^x.27^{2.x}=\left(-9\right)^4\)
e,\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
g,\(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)
h,\(|3x-5|=|\frac{1}{2}x+3|\)
i,\(\left(\frac{3}{4}-x\right)^3=\frac{1}{64}\)
Tìm x:
|5x-3|-3x=7
|x-3|+|x-5|-4x=-28
\(\left|x+2\right|+\left|x+\frac{3}{5}\right|+\left|x+\frac{1}{2}\right|=4x\)
\(\left|2x-1\right|+\left(4x^2-1\right)^2=0\)