(x-2)2=(x-2)5
(cho xin lời giải)
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H
30 tháng 9 2023
`a, 3/4 - 5/4 :(x-1) =1/2`
`=> 5/4:(x-1)= 3/4 -1/2`
`=> 5/4:(x-1)= 3/4 - 2/4`
`=> 5/4:(x-1)= 1/4`
`=> x-1= 5/4 : 1/4`
`=> x-1=5`
`=>x=5+1`
`=>x=6`
__
`(1/2-x)^2 -2^2 =12`
`=> (1/2-x)^2 = 12+4`
`=> (1/2-x)^2= 16`
`=> (1/2-x)^2 =4^2`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=4\\\dfrac{1}{2}-x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\)
__
`(1/2)^(2x-1) =1/16`
`=> (1/2)^(2x-1) = (1/2)^4`
`=> 2x-1=4`
`=> 2x=4+1`
`=>2x=5`
`=>x=5/2`
VQ
30 tháng 9 2023
\(a,\dfrac{3}{4}-\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\dfrac{5}{4}:\left(x-1\right)=\dfrac{1}{4}\)
\(x-1=\dfrac{5}{4}:\dfrac{1}{4}\)
\(x-1=5\)
\(x=6\)
\(\left(\dfrac{1}{2}-x\right)^2-2^2=12\)
\(\left(\dfrac{1}{2}-x\right)^2-4=12\)
\(\left(\dfrac{1}{2}-x\right)^2=16\)
\(\left(\dfrac{1}{2}-x\right)^2=4^2hoặc\left(\dfrac{1}{2}-x\right)^2=\left(-4\right)^2\)
\(\dfrac{1}{2}-x=4hoặc\dfrac{1}{2}-x=-4\)
=>1/2 -x =4 1/2 -x= -4
=> x=1/2-4 x=1/2-(-4)
=>x=-7/2 x=9/2
vậy x∈{-7/2 ; 9/2}
\(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{16}\)
\(=>\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^4\)
\(=>2x-1=4\)
\(=>2x=5\)
\(=>x=\dfrac{5}{2}\)
TL
6
TA
4 tháng 7 2023
\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+2\sqrt{2\left(x-2\right)}}+\sqrt{x-2\sqrt{2\left(x-2\right)}}=2\sqrt{2}\)
\(\Leftrightarrow2x+2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8\)
\(\Leftrightarrow2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8-2x\)
\(\Leftrightarrow4\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]=64-32x+4x^2\)
\(\Leftrightarrow4x^2-32x+64=64-32x+4x^2+\)
\(\Leftrightarrow64=64\) (Đúng)
⇒ Phương trình có vô số nghiệm.
Vậy \(S=\mathbb R\).
3V
3
O
18 tháng 3 2022
\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dfrac{6}{7}\times\dfrac{7}{8}\times\dfrac{8}{9}\)
\(=\dfrac{1\times2\times3\times4\times5\times6\times7\times8}{2\times3\times4\times5\times6\times7\times8\times9}\)
\(=\dfrac{1}{9}\)
2
3 tháng 7 2023
`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}` `(1 < x < 2)`
`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`
`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`
`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]` (Vì `1 < x < 2`)
`=3/2x - 3/2 + 3/2 + 5/2x`
`=4x`
Y
28 tháng 6 2023
\(\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\) có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
20 tháng 12 2021
c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)