Q = (1 - \(\dfrac{1}{\sqrt{x}}\))\(^2\):(\(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\))
Các bạn giúp mình với
Những câu hỏi liên quan
Rút gọn các biểu thức sau:Cleft(dfrac{sqrt{a}}{sqrt{a}-1}-dfrac{sqrt{a}}{a-sqrt{1}}right):dfrac{sqrt{a}+1}{a-1}Dleft(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x}+2}right).dfrac{sqrt{x}+1}{sqrt{x}-1}Eleft(1+dfrac{x+sqrt{x}}{sqrt{x}+1}right)left(1+dfrac{x-sqrt{x}}{1-sqrt{x}}right)làm ơn giúp mình với ạ!!mình đang cần gấp vào tối nay ạ!!
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Rút gọn các biểu thức sau:
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
làm ơn giúp mình với ạ!!mình đang cần gấp vào tối nay ạ!!
\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(E=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
ĐK : a >= 0 , a khác 1
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)
giúp mình câu rút gọn với ạ :3
\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}\)
\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1)))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`
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\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1))))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`
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Giúp mình với!Cho biểu thức: C dfrac{3x+sqrt{9x}-3}{x+sqrt{x}-2}-dfrac{sqrt{x}+1}{sqrt{x}+2}+dfrac{sqrt{x}-2}{sqrt{x}}timesleft(dfrac{1}{1-sqrt{x}}-1right)a) Rút gọn C.b) Tìm các giá trị của x để C sqrt{x}c) Tìm giá trị của C, biết |2x - 5| 3.d) So sánh C và C^2
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Giúp mình với!
Cho biểu thức: C = \(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\times\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
a) Rút gọn C.
b) Tìm các giá trị của x để C = \(\sqrt{x}\)
c) Tìm giá trị của C, biết |2x - 5| = 3.
d) So sánh C và \(C^2\)
a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)
=>\(x-2\sqrt{x}-1=0\)
=>\(\Leftrightarrow x=3+2\sqrt{2}\)
c: |2x-5|=3
=>2x-5=3 hoặc 2x-5=-3
=>2x=2 hoặc 2x=8
=>x=4(nhận) hoặc x=1(loại)
Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)
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left(dfrac{2x+1}{sqrt{x^3-1}}-dfrac{sqrt{x}}{x+sqrt{x+1}}right)left(dfrac{1+sqrt{x^3}}{1+sqrt{x}}-sqrt{x}right)
dfrac{sqrt{x-2sqrt{x-1}+}sqrt{x+2sqrt{x-1}}}{sqrt{dfrac{1}{x^2}-dfrac{2}{x}+1}}
left(dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-sqrt{xy}right):left(x-yright)+dfrac{2sqrt{y}}{sqrt{x}+sqrt{y}}
left(dfrac{sqrt{x}}{3+sqrt{x}}+dfrac{x+9}{9-x}right):left(dfrac{3sqrt{x}+1}{x-3sqrt{x}}-dfrac{1}{sqrt{x}}right)
Mí bạn giải giúp mik bài này nhé!:))))))))))))))))))))))))))))))ARIGATO MÍ BẠN NHI...
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\(\left(\dfrac{2x+1}{\sqrt{x^3-1}}-\dfrac{\sqrt{x}}{x+\sqrt{x+1}}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(\dfrac{\sqrt{x-2\sqrt{x-1}+}\sqrt{x+2\sqrt{x-1}}}{\sqrt{\dfrac{1}{x^2}-\dfrac{2}{x}+1}}\)
\(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
Mí bạn giải giúp mik bài này nhé!:))))))))))))))))))))))))))))))ARIGATO MÍ BẠN NHIW!!!!!!!!!!:)))))))))))))))))))))))))))))))))
\(a.\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)=\dfrac{x+1+\sqrt{x}}{x\sqrt{x}-1}.\dfrac{x\sqrt{x}+1-\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
\(b.ĐK:x>2\) ( thường là những bài rút gọn sẽ kèm theo ĐK nhé , mình thêm như vậy , nếu không bạn chia TH ra )
\(\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{\dfrac{1}{x^2}-\dfrac{2}{x}+1}}=\dfrac{\sqrt{x-1}-1+\sqrt{x-1}+1}{1-\dfrac{1}{x}}=\dfrac{2\sqrt{x-1}}{1-\dfrac{1}{x}}\)
\(c.\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=1\)
\(d.Tuong-tự\)
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\(\left\{\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right\}:\left\{1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right\}\)
a, Rút gọn P
b, Tìm các giá trị của x để P = \(\dfrac{6}{5}\)
Giúp mình với ạ, Cảm ơn trước!
\(a,\)
\(=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right):\left(\dfrac{3}{3\sqrt{x}+1}\right)\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+3x}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\dfrac{3\sqrt{x}+1}{3}\)
\(=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}\)
\(b,\)Thay \(P=\dfrac{6}{5}\) vào pt, ta có :
\(\dfrac{3\sqrt{x}+1}{3\sqrt{x}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(3\sqrt{x}+1\right)=6\left(3\sqrt{x}-1\right)\)
\(\Leftrightarrow15\sqrt{x}+5-18\sqrt{x}+6=0\)
\(\Leftrightarrow-3\sqrt{x}+11=0\)
\(\Leftrightarrow-3\sqrt{x}=-11\)
\(\Leftrightarrow\sqrt{x}=\dfrac{11}{3}\)
\(\Leftrightarrow x=\left(\dfrac{11}{3}\right)^2\)
\(\Leftrightarrow x=\dfrac{121}{9}\)
Vậy \(x=\dfrac{121}{9}\) thì \(P=\dfrac{6}{5}\)
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rút gọn
dfrac{9-x}{sqrt{x}+3}-dfrac{9-6sqrt{x}+x}{sqrt{x}-3}-6 (với x_9)
left(dfrac{2sqrt{x}}{xsqrt{x}+x+sqrt{x}+1}-dfrac{1}{sqrt{x}+1}right)/left(dfrac{2sqrt{x}}{sqrt{x}+1}-1right) (với x0, x#1)
sqrt{x+12+6sqrt{x+3}}-sqrt{x+12-6sqrt{x+3}} ( với x_6)
sqrt{m^2+6m+9}+sqrt{m^2-6m+9} (m bát kì)
dfrac{xsqrt{x}-1}{x-sqrt{x}}-dfrac{xsqrt{x}+1}{x+sqrt{x}}dfrac{x+1}{sqrt{x}}
dfrac{xsqrt{y}+ysqrt{x}}{sqrt{xy}}/dfrac{sqrt{x}-sqrt{y}}{x-y}
left(dfrac{1}{sqrt{x}-1}+dfrac{1}{sqrt{x}+1}right)left...
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rút gọn
\(\dfrac{9-x}{\sqrt{x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\) (với x>_9)
\(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)/\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) (với x>=0, x#1)
\(\sqrt{x+12+6\sqrt{x+3}}-\sqrt{x+12-6\sqrt{x+3}}\) ( với x>_6)
\(\sqrt{m^2+6m+9}+\sqrt{m^2-6m+9}\) (m bát kì)
\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\dfrac{x+1}{\sqrt{x}}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}/\dfrac{\sqrt{x}-\sqrt{y}}{x-y}\)
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(\left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right)/\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}\)
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
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cho A=\(\dfrac{1}{\sqrt{x}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x\sqrt{x-x}}{1-\sqrt{x}}\)
rút gọn biểu thức A
giúp mình với mình cảm ơn
\(A=\dfrac{\sqrt{x}-\sqrt{x-1}-\sqrt{x}-\sqrt{x-1}}{x-x+1}=-2\sqrt{x-1}\)
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30 tháng 8 2021 lúc 17:06
Thu gọn giúp với ak
a) \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)
b) \(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right).\left(x-3\sqrt{x}+2\right)\)
a:Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
b: Ta có: \(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(x-3\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
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26 tháng 7 2021 lúc 16:37
có thể giúp mình giải bài này với đc k ạ mình đang cần gấp (xin cảm ơn)Bài 1:a,3x-7sqrt{x}+40b, dfrac{1}{2}sqrt{x-1}-dfrac{9}{2}sqrt{x-1}+3sqrt{x-1}-17c, 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}-1Bài 2:a, sqrt{x^2+6x+9}3x-6b, sqrt{3x^2}x+2c, sqrt{x^2-4x+4}-2x+50d, x^2-2sqrt{7x}+70Bài 3:a, sqrt{3+x}+sqrt{6-x}3b, sqrt{3+x}-sqrt{2-x}1
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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}}`
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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}`
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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
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