G=\(\frac{3\sqrt{x}}{\sqrt{x}-3}\) =\(\left(\frac{3\sqrt{x}}{\sqrt{x}-3}\right)^{^4}\) với x≥ 0, x≠9 Tìm x
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Tìm ĐKXĐ và rút gọn biểu thứcAfrac{left(sqrt{a}+sqrt{b}right)^2-4sqrt{ab}}{sqrt{a}-sqrt{b}}-frac{asqrt{b}+bsqrt{a}}{sqrt{ab}}Bleft(frac{2sqrt{x}-x}{xsqrt{x}-1}-frac{1}{sqrt{x}-1}right):frac{x-1}{x+sqrt{x}+1}Cleft(1-frac{x-3sqrt{x}}{x-9}right):left(frac{sqrt{x}-3}{2-sqrt{x}}+frac{sqrt{x}-2}{3+sqrt{x}}-frac{9-x}{x+sqrt{x}-6}right)Dleft(frac{sqrt{x}}{3+sqrt{x}}+frac{x+9}{9-x}right):left(frac{3sqrt{x}+1}{x-3sqrt{x}}-frac{1}{sqrt{x}}right)CM rằng GT của bthức A ko phụ thuộc vào aTìm x để C 4Tìm x sa...
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Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
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RÚT GỌN BIỂU THỨC Afrac{a-b}{sqrt{a}-sqrt{b}}-frac{sqrt{a^3}-sqrt{b^3}}{a-b}(với a_ 0, b_ 0, a#b)Bleft(frac{sqrt{x^3}+sqrt{y^3}}{sqrt{x}+sqrt{y}}-sqrt{xy}right).left(frac{sqrt{x}+sqrt{y}}{x-y}right)(với x_ 0, y_ 0, x#y)Cx-4-sqrt{16-8x^2+x^4}(với x4)Dfrac{a+b-2sqrt{ab}}{sqrt{a}-sqrt{b}}:frac{1}{sqrt{a}+sqrt{b}}(với a0, b0, a#b)Eleft(2+frac{a-sqrt{a}}{sqrt{a}-1}right).left(2-frac{a+sqrt{a}}{sqrt{a}+1}right)(với a0, a#1)Ffrac{a-3sqrt{a}}{sqrt{a}-3}-frac{a+4sqrt{a}+3}{sqrt{a}+3}( với a_ 9...
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RÚT GỌN BIỂU THỨC
A=\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)(với a>_ 0, b>_ 0, a#b)
B=\(\left(\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right).\left(\frac{\sqrt{x}+\sqrt{y}}{x-y}\right)\)(với x>_ 0, y>_ 0, x#y)
C=\(x-4-\sqrt{16-8x^2+x^4}\)(với x>4)
D=\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}\)(với a>0, b>0, a#b)
E=\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right).\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\)(với a>0, a#1)
F=\(\frac{a-3\sqrt{a}}{\sqrt{a}-3}-\frac{a+4\sqrt{a}+3}{\sqrt{a}+3}\)( với a>_ 9)
G=\(\frac{9-x}{\sqrt{x}+3}-\frac{9-6\sqrt{x}+x}{\sqrt{x}-3}-6\)( với x>_ 9 )
Cho biểu thức P=\(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
với x>=0 ; x khác 9; x khác 4
Rút gọn biểu thức P
giúp mình với
Bài làm:
Ta có:
\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)
\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)
Mọi người cho em hỏi:left(frac{sqrt{x}}{left[sqrt{x}-3right]left[sqrt{x}+3right]}+frac{2}{sqrt{x}+3}-frac{3}{sqrt{x}-3}right):left(sqrt{x}-3+frac{12-x}{sqrt{x}+3}right)left(frac{sqrt{x}+2left[sqrt{x}-3right]-3left[sqrt{x}+3right]}{left[sqrt{x}-3right]left[sqrt{x}+3right]}right):left(frac{left[sqrt{x}-3right]left[sqrt{x}+3right]+12-x}{sqrt{x}-3}right)left(frac{sqrt{x}+2sqrt{x}-6-3sqrt{x}-9}{left[sqrt{x}+3right]left[sqrt{x}-3right]}right):left(frac{x+3sqrt{x}-3sqrt{x}-9+12-x}{sqrt{x}+3}right)left(...
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Mọi người cho em hỏi:
\(\left(\frac{\sqrt{x}}{\left[\sqrt{x}-3\right]\left[\sqrt{x}+3\right]}+\frac{2}{\sqrt{x}+3}-\frac{3}{\sqrt{x}-3}\right):\left(\sqrt{x}-3+\frac{12-x}{\sqrt{x}+3}\right)\)
\(\left(\frac{\sqrt{x}+2\left[\sqrt{x}-3\right]-3\left[\sqrt{x}+3\right]}{\left[\sqrt{x}-3\right]\left[\sqrt{x}+3\right]}\right):\left(\frac{\left[\sqrt{x}-3\right]\left[\sqrt{x}+3\right]+12-x}{\sqrt{x}-3}\right)\)
\(\left(\frac{\sqrt{x}+2\sqrt{x}-6-3\sqrt{x}-9}{\left[\sqrt{x}+3\right]\left[\sqrt{x}-3\right]}\right):\left(\frac{x+3\sqrt{x}-3\sqrt{x}-9+12-x}{\sqrt{x}+3}\right)\)
\(\left(\frac{-15}{\left[\sqrt{x}+3\right]\left[\sqrt{x}-3\right]}\right):\left(\frac{3}{\sqrt{x}+3}\right)\)
\(\left(\frac{-15}{\left[\sqrt{x}+3\right]\left[\sqrt{x}-3\right]}\right).\left(\frac{\sqrt{x}+3}{3}\right)\)
\(\frac{-5}{\sqrt{x}-3}\)
\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right)\):\(\left(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)rút gọn và tìm gtln
Mình tách thành hai phần nhìn cho dễ hiểu nhé !
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
+) \(\frac{x-3\sqrt{x}}{x-9}-1=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}-1=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}\)
+) \(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{9-x+x-9-x+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)
\(=\frac{3\left(\sqrt{x}-2\right)}{x-4}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)
Cho biểu thức: \(P=\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}+2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right):\left(1-\frac{3\sqrt{x}-9}{x-9}\right)\)
a)Rút gọn biểu thức
b)Tính P với \(x=\frac{\sqrt{4+2\sqrt{3}}\left(\sqrt{x}-1\right)}{\sqrt{6+2\sqrt{5}-\sqrt{5}}}\)
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
1/ Cho Dfrac{x^2-sqrt{x}}{x+sqrt{x}+1}-frac{x^2+sqrt{x}}{x-sqrt{x}+1}với 0≤x≤1
a) Rút gọn
b) CMinh 1-sqrt{D+x+1}sqrt{x}
2/Cho Efrac{15sqrt{x}-11}{x+2sqrt{x}-3}+frac{3sqrt{x}-2}{1-sqrt{x}}-frac{2sqrt{x}+3}{sqrt{x}+3}với x≥0 và x≠1
a) Rút gọn
b) Tìm giá trị của x để E frac{1}{2}
c) So sánh E với frac{2}{3}
3/Cho Gleft(frac{x-3sqrt{x}}{x-9}-1right):left(frac{9-x}{x+sqrt{x}-6}+frac{sqrt{x}-3}{sqrt{x}-2}-frac{sqrt{x}+2}{sqrt{x}+3}right)với x≥0,x≠4,x≠9
a) Rút gọn
b) Tìm x để G1
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1/ Cho D=\(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}\)với 0≤x≤1
a) Rút gọn
b) CMinh 1\(-\sqrt{D+x+1}=\sqrt{x}\)
2/Cho E=\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)với x≥0 và x≠1
a) Rút gọn
b) Tìm giá trị của x để E = \(\frac{1}{2}\)
c) So sánh E với \(\frac{2}{3}\)
3/Cho G=\(\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)với x≥0,x≠4,x≠9
a) Rút gọn
b) Tìm x để G<1
Rút gọn \(\left(\frac{\sqrt{X}}{3+\sqrt{X}}+\frac{X+9}{9-X}\right):\frac{X-3\sqrt{X}}{2\sqrt{X}+4}\left(X>0,X\ne9\right)\)
Rút gọn:a/ frac{left(sqrt{x^2+9}-3right)left(sqrt{x^2+9}+3right)left(x+sqrt{xy}+yright)sqrt{x-2sqrt{xy}+y}}{xleft(xsqrt{x}-ysqrt{y}right)} (với x0, yge0, xney b/ left[left(frac{1}{sqrt{x}}+frac{1}{sqrt{y}}right).frac{2}{sqrt{x}+sqrt{y}}+frac{1}{sqrt{x}}+frac{1}{sqrt{y}}right]:frac{sqrt{x^3}+ysqrt{x}+xsqrt{y}+sqrt{y^3}}{sqrt{x^3y}+sqrt{xy^3}}(với x0 và xne1 c/ left(frac{sqrt{x}+1}{sqrt{xy}+1}+frac{sqrt{xy}+sqrt{x}}{1-sqrt{xy}}+1right):left(1-frac{sqrt{xy}+sqrt{x}}{sqrt{xy}-1}-frac{sqrt{x}+1}{sqrt...
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Rút gọn:
a/ \(\frac{\left(\sqrt{x^2+9}-3\right)\left(\sqrt{x^2+9}+3\right)\left(x+\sqrt{xy}+y\right)\sqrt{x-2\sqrt{xy}+y}}{x\left(x\sqrt{x}-y\sqrt{y}\right)}\) (với x>0, y\(\ge\)0, x\(\ne\)y
b/ \(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right]:\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)(với x>0 và x\(\ne\)1
c/ \(\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)(với x>0 và x\(\ne\)1