tìm X biết: 2.|1/2.X-1/3|-3/2=1/4
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\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(\Leftrightarrow3x-6+4x-4=25\)
\(\Leftrightarrow7x=35\)
\(\Leftrightarrow x=5\)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
a) \(2^x=8\)
⇔ \(2^x=2^3\)
⇒ \(x=3\)
b) \(3^x=27\)
⇔ \(3^x=3^3\)
⇒ \(x=3\)
c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)
d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)
d) \(\left(x+1\right)^3=-125\)
⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)
⇔ \(x+1=-5\)
⇔ \(x=-5-1=-6\)
2:
a: (x-1,2)^2=4
=>x-1,2=2 hoặc x-1,2=-2
=>x=3,2(loại) hoặc x=-0,8(loại)
b: (x-1,5)^2=9
=>x-1,5=3 hoặc x-1,5=-3
=>x=-1,5(loại) hoặc x=4,5(loại)
c: (x-2)^3=64
=>(x-2)^3=4^3
=>x-2=4
=>x=6(nhận)
Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
\(2\cdot\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(=>2\cdot\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(=>\left|\frac{1}{2}\cdot x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{8}\)
\(=>\frac{1}{2}\cdot x-\frac{1}{3}=\text{±}\frac{7}{8}\)
\(+\frac{1}{2}\cdot x-\frac{1}{3}=\frac{7}{8}\)
\(=>\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}\)
\(=>x=\frac{29}{24}:\frac{1}{2}=\frac{29}{12}\)
\(+\frac{1}{2}\cdot x-\frac{1}{3}=-\frac{7}{8}\)
\(=>\frac{1}{2}\cdot x=-\frac{7}{8}+\frac{1}{3}=-\frac{13}{24}\)
\(=>x=-\frac{13}{24}:\frac{1}{20}=-\frac{13}{12}\)
\(2.\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2.\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}\)
\(2.\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{8}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}\\\frac{1}{2}x=\frac{-7}{8}+\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=\frac{29}{24}\\\frac{1}{2}x=\frac{-13}{24}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{29}{24}:\frac{1}{2}\\x=\frac{-13}{24}:\frac{1}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{29}{12}\\x=\frac{-13}{12}\end{cases}}}\)
Vậy .....................................
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