a/ (x - 1)6 = (x - 1)8
=> (x - 1)6 [1 - (x - 1)2] = 0
=> (x - 1)6 (1 - x2 + 2x - 1) = 0
=> (x - 1)6 (-x2 + 2x) = 0
=> x - 1 = 0 => x = 1
hoặc - x2 + 2x = 0 => x = 0 hoặc x = 2
Vậy x = 0, x = 1, x = 2
Tìm x biết:
a) \(\left|x+2\dfrac{1}{2}\right|=\left|3x+1\right|\)
b) \(\left|2x-6\right|+\left|x+3\right|=8\)
c) \(2.\left|x+2\right|+\left|4-x\right|=11\)
Tìm x và y biết:
d)\(-1\frac{2}{3}-\left(\left|2x\right|+\frac{5}{6}\right)=\)\(-2\)e)\(\left(-\frac{1}{2}+\frac{1}{3}\right):\left|1-2x\right|-1\frac{1}{4}:\left(-\frac{5}{8}\right).\left(-\frac{1}{2}\right)^2=\frac{1}{3}\)
c)\(\left|2x-1\right|+\left|2y+1\right|+\left|2x-y\right|=0\)b)\(\left|2x-1\right|=2x-1\)
a)\(\left|x-3\right|=x+4\)
Bài 1 : Tìm GTNN của : \(A=\left|x+8\right|+\left|2x+7\right|+\left|3x+6\right|+\left|4x-7\right|+\left|3x-6\right|+\left|2x-7\right|+\left|x-8\right|-100\)
Bài 4 : Tìm \(x\), biết :
a ) \(\left|4-2x\right|+\left|x+3\right|=8\)
b ) \(\left|x+1\right|+\left|x-2\right|+\left|x+3\right|=6\)
c ) \(\left|x-2\right|+\left|x-3\right|+\left|4-x\right|=2\)
d ) \(2\left|x+2\right|+\left|4-3x\right|=11\)
e ) \(\left|2x-1\right|+\left|2x-5\right|=4\)
g ) \(\left|x-3\right|+\left|3x+4\right|=\left|2x+1\right|\)
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
Bài 1 : Tìm GTNN của : \(A=\left|x+8\right|+\left|2x+7\right|+\left|3x+6\right|+\left|4x-7\right|+\left|3x-6\right|+\left|2x-7\right|+\left|x-8\right|-100\)
\(a)\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)
\(b)\left(x+\frac{1}{2}\right)^2=\frac{4}{9}\)
\(_{c)\left(2x-3\right)^6=\left(2x-3\right)^6}\)
Giải đầy đủ nhé!
TÍNH
a.\(-\dfrac{5}{4}x^4.\dfrac{8}{15}x\) b.\(-2x\left(\dfrac{3}{4}x^2-x+\dfrac{1}{2}\right)\) c.\(x\left(x-\dfrac{1}{2}\right)-\left(x-2\right)\left(x+3\right)\)
