\(\sqrt{\dfrac{x^2+1}{1-2x}}\)
giúp mình vs mình đang cần gấp
ai có thể giúp mình giải bài này vs đc không mình đang cần rất gấp (làm chi tiết hộ mình nhé, xin cảm ơn)
Bài 4:
a, \(\sqrt{3x+4}-\sqrt{2x+1}=\sqrt{x+3}\)
b, \(\sqrt{2x-5}+\sqrt{x+2}=\sqrt{2x+1}\)
c, \(\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
d,\(\sqrt{x+9}=5-\sqrt{2x+4}\)
Bài 5:
a, \(\sqrt{x+4\sqrt{x}+4}=5x+2\)
b, \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)
c, \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
d,\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)
Ví Dụ 1:
a, \(\sqrt{2x-1}=\sqrt{2}-1\)
b, \(\sqrt{x+5}=3-\sqrt{2}\)
c, \(\sqrt{3x^2}-\sqrt{12}=0\)
d, \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
Vd1:
d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)
\(\Leftrightarrow x=6\)
Giúp mình câu này với mik đang cần gấp
\(\dfrac{2x^2-x}{x-1}\)+\(\dfrac{x+1}{1-x}\)+\(\dfrac{2-x^2}{x-1}\)
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
Giúp mình với, mình đang cần gấp
\(\dfrac{x^3}{1-x}\) + x2 + x + 1
\(\dfrac{2x}{x^2+4x+4}\) + \(\dfrac{x+1}{x+2}\) + \(\dfrac{2-x}{\left(x+2\right)^2}\)
Xin cảm ơn <3
\(a,=\dfrac{x^3-\left(x-1\right)\left(x^2+x+1\right)}{1-x}=\dfrac{x^3-x^3+1}{1-x}=\dfrac{1}{1-x}\\ b,=\dfrac{2x+x^2+3x+2+2-x}{\left(x+2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x+2\right)^2}=1\)
3)tìm GTLN của \(A=\dfrac{\sqrt{x-9}}{5x}\)
giúp mình vs mk đang cần gấp
\(A=\dfrac{\sqrt{x-9}}{5x}\left(ĐKx\ge9\right)\)
A'=\(\dfrac{\dfrac{5x}{2\sqrt{x-9}}-5\sqrt{x-9}}{\left(5x^2\right)}\)
\(A'=0\rightarrow5x=10\left(x-9\right)\)
\(\rightarrow x=18\)
\(MaxA=\dfrac{1}{30}\) khi \(x=18\)
\(A=\dfrac{2.3\sqrt{x-9}}{30x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)
\(A_{max}=\dfrac{1}{30}\) khi \(\sqrt{x-9}=3\Leftrightarrow x=18\)
có thể giúp mình giải bài này với đc k ạ mình đang cần gấp (xin cảm ơn)
Bài 1:
a,\(3x-7\sqrt{x}+4=0\)
b, \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
c, \(\dfrac{\sqrt{x}-2}{\sqrt{x}-4}=\dfrac{6-\sqrt{x}}{7-\sqrt{x}}\)
d, \(\sqrt{x-3}-\dfrac{5}{3}\sqrt{9x-27}+\dfrac{3}{2}\sqrt{4x-12}=-1\)
Bài 2:
a, \(\sqrt{x^2+6x+9}=3x-6\)
b, \(\sqrt{3x^2}=x+2\)
c, \(\sqrt{x^2-4x+4}-2x+5=0\)
d, \(x^2-2\sqrt{7x}+7=0\)
Bài 3:
a, \(\sqrt{3+x}+\sqrt{6-x}=3\)
b, \(\sqrt{3+x}-\sqrt{2-x}=1\)
Bài 2
b, `\sqrt{3x^2}=x+2` ĐKXĐ : `x>=0`
`=>(\sqrt{3x^2})^2=(x+2)^2`
`=>3x^2=x^2+4x+4`
`=>3x^2-x^2-4x-4=0`
`=>2x^2-4x-4=0`
`=>x^2-2x-2=0`
`=>(x^2-2x+1)-3=0`
`=>(x-1)^2=3`
`=>(x-1)^2=(\pm \sqrt{3})^2`
`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$
Vậy `S={1+\sqrt{3};1-\sqrt{3}}`
Bài 1
a, `3x-7\sqrt{x}+4=0` ĐKXĐ : `x>=0`
`<=>3x-3\sqrt{x}-4\sqrt{x}+4=0`
`<=>3\sqrt{x}(\sqrt{x}-1)-4(\sqrt{x}-1)=0`
`<=>(3\sqrt{x}-4)(\sqrt{x}-1)=0`
TH1 :
`3\sqrt{x}-4=0`
`<=>\sqrt{x}=4/3`
`<=>x=16/9` ( tm )
TH2
`\sqrt{x}-1=0`
`<=>\sqrt{x}=1` (tm)
Vậy `S={16/9;1}`
b, `1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17` ĐKXĐ : `x>=1`
`<=>(1/2-9/2+3)\sqrt{x-1}=-17`
`<=>-\sqrt{x-1}=-17`
`<=>\sqrt{x-1}=17`
`<=>x-1=289`
`<=>x=290` ( tm )
Vậy `S={290}`
Bài 1:
a) Ta có: \(3x-7\sqrt{x}+4=0\)
\(\Leftrightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
a)\(\sqrt{-3x+5}=7\)
b)\(\sqrt{\dfrac{1}{2}x+6}=\sqrt{2}\)
c)\(\sqrt{2x+1}=1+2\sqrt{3}\)
d)\(\sqrt{\left(3x-1\right)^2}=8\)
giúp mình với ạ mình cần gấp
\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(\left(2-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(2-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
mọi người giúp mình với ạ mình đang cần gấp mình cảm ơn
Bài 2.4 Chứng minh với mọi giá trị của x để biểu thức có nghĩa thì giá trị của:
𝐴= \((\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x+3}}{2\sqrt{x}+2}).\dfrac{4x-4}{5}\) không phụ thuộc vào x.
ai bt giúp mình với mình đang cần gấp
A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)
A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)
A = 4
Vậy A không phụ thuộc vào x
Chúc bn học tốt!
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)
\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)
\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)
\(=10\cdot\dfrac{2}{5}=4\)
giúp mình với ạ , mình đang cần gấp!!!
(\(\dfrac{1}{2}\)+2x).(2x-3)=0
=>2x+1/2=0 hoặc 2x-3=0
=>x=-1/4 hoặc x=3/2
\(\left[{}\begin{matrix}\dfrac{1}{2}+2x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
( 1/2 + 2x ) ( 2x - 3 ) = 0
( 1/2 + 2x ) = 0
( 2x - 3 ) = 0
( 1/2 + 2x ) = 0
\(\Leftrightarrow\) ( 2x + 1/2 ) = 0
2x + 1/2 - 1/2 = 0 - 1/2
2x = -1/2
2x/2 = -1/2 : 2
x = -1/4
2x - 3 = 0
2x - 3 + 3 = 0 + 3
2x = 3
2x/2 = 3/2
x = 3/2
\(\Rightarrow\) x1 = -1/4 ; x2 = 3/2