cho E = \(\left(\dfrac{3+\sqrt{x}}{x-1}+\dfrac{3}{\sqrt{x}+1}\right):\dfrac{4}{x+\sqrt{x}}\)
a. Tìm ĐK và rút gọn E
b. Tính E khi x = \(\dfrac{9}{4}\)
c. Tìm x để E < 0
Bài 2:
Cho biểu thức E= \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
a) Rút gọn E
b) Tìm x để E= 2
c) Tính giá trị của E khi x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
a: Ta có: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+4\sqrt{x}\right):\left(\dfrac{x-1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4x^2}{\left(x-1\right)^2}\)
b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-2x^2+4x-2=0\)
\(\Leftrightarrow2x^2+4x-2=0\)
\(\Leftrightarrow x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}-1\\x=\sqrt{2}-1\end{matrix}\right.\)
c: Ta có: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=2\)
Thay x=2 vào E, ta được:
\(E=\dfrac{4\cdot2^2}{1}=16\)
E = \(\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right)\): \(\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\right)\)với x > 0, x ≠ 4 và x ≠ 9
a) Rút gọn biểu thức E
b) Tính x để E dương
\(a)E=\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-1\right):\left(\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+2}\right):\dfrac{4-x+x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{9-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2}{\sqrt{x}+2}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{9-x}\\ =\dfrac{-2\left(\sqrt{x}-3\right)}{9-x}=\dfrac{2\left(\sqrt{x}-3\right)}{x-9}\\ =\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2}{\sqrt{x}+3}\)
\(b)\)E dương
\(\Leftrightarrow E>0\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}>0\\ \Leftrightarrow\sqrt{x}+3>0\left(Vì.2>0\right)\\ \Leftrightarrow\sqrt{x}>-3\forall x\in R\\ \Rightarrow x\ge0\)
Kết hợp đk
\(x\ge0;x\ne4;x\ne9\)
Vậy \(x\ge0;x\ne4;x\ne9\) thì E dương
1`,\(E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)(x>0,x\(\ne\)1)
a,rút gọn E b,Tìm x để E > 0
2,\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right).\left(\sqrt{x}+1\right)\) (x>0,x≠1)
a,rút gọn B b,tìm x để G=2
\(1,\\ a,E=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>1\\ 2,\\ a,B=\dfrac{x-\sqrt{x}+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,B=2\Leftrightarrow\sqrt{x}-1=2\left(\sqrt{x}+1\right)\\ \Leftrightarrow\sqrt{x}-1=2\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}=-3\Leftrightarrow x\in\varnothing\)
Cho biểu thức:
E = (\(\dfrac{1}{x+\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x}+1}\)) : \(\dfrac{2}{\sqrt{x}-2}\)
a) Rút gọn E
b) Tính giá trị E khi x = 19 - \(8\sqrt{3}\)
c) tìm x để E = -1
d) Tìm x để E = \(\dfrac{1}{\sqrt{x}}\)
e) Tìm x để E > 0
f) So sánh E với \(\dfrac{1}{2}\)
g) Tìm x \(\in\) Z để \(\dfrac{1}{E}\)\(\in\) Z
h) Với x > 4. So sánh: E và \(\sqrt{E}\)
\(a,ĐK:x>0;x\ne4\\ E=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}-2}{2\sqrt{x}}\\ b,x=19-8\sqrt{3}=\left(4-\sqrt{3}\right)^2\\ \Leftrightarrow E=\dfrac{4-\sqrt{3}-2}{2\left(4-\sqrt{3}\right)}=\dfrac{\left(2-\sqrt{3}\right)\left(4+\sqrt{3}\right)}{26}=\dfrac{5-2\sqrt{3}}{26}\\ c,E=-1\Leftrightarrow\sqrt{x}-2=-2\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{4}{9}\left(tm\right)\\ d,E=\dfrac{1}{\sqrt{x}}\Leftrightarrow\dfrac{\sqrt{x}-2}{2}=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)
\(e,E>0\Leftrightarrow\sqrt{x}-2>0\left(2\sqrt{x}>0\right)\Leftrightarrow x>4\\ f,E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}=\dfrac{1}{2}-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\left(-\dfrac{1}{\sqrt{x}}< 0\right)\\ g,\dfrac{1}{E}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(4\right)=\left\{-1;0;1;2;4\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;2;3;4;6\right\}\\ \Leftrightarrow x\in\left\{1;9;16;36\right\}\left(x\ne4\right)\\ h,x>4\Leftrightarrow\sqrt{x}-2>0\\ \Leftrightarrow E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}>0\Leftrightarrow E\ge\sqrt{E}\)
\(C=\left(\dfrac{2\sqrt{x}}{2x-5\sqrt{x}+3}-\dfrac{5}{2\sqrt{x}-3}\right)\div\left(3+\dfrac{2}{1-\sqrt{x}}\right)\)
a) Rút gọn C
b) Tính C với \(x=\dfrac{2}{2-\sqrt{3}}\)
c) Tìm x để C= –1
d) Tìm x để C > 0
e) So sánh C’ với –2
f) Tìm GRNN của C’ với C’=\(\dfrac{1}{C}\times\dfrac{1}{\sqrt{x}+1}\)
i)Tìm \(x\in Z\) để \(C'\in Z\) g) Tìm m để pt C’.m = –1 có nghiệm
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)
a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)
\(=\dfrac{2\sqrt{x}-5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-3-2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=-\dfrac{1}{2\sqrt{x}-3}\)
b: \(x=\dfrac{2}{2-\sqrt{3}}=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)
Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)
c: C=-1
=>\(2\sqrt{x}-3=1\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
d: C>0
=>\(2\sqrt{x}-3< 0\)
=>\(\sqrt{x}< \dfrac{3}{2}\)
=>\(0< =x< \dfrac{9}{4}\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)
Cho E = \(\dfrac{\sqrt{x}}{x+2\sqrt{x}}\): \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
a) Rút gọn biểu thức E
b) Tìm x khi E = \(\dfrac{2}{5}\)
a: ĐKXĐ: x>0
\(E=\dfrac{\sqrt{x}}{x+2\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\dfrac{1}{\sqrt{x}+2}:\dfrac{\sqrt{x}+2+x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x+\sqrt{x}+2}\)
b: E=2/5
=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{2}{5}\)
=>\(5\sqrt{x}=2x+2\sqrt{x}+4\)
=>\(2x-3\sqrt{x}+4=0\)
=>\(x-\dfrac{3}{2}\cdot\sqrt{x}+2=0\)
=>\(x-2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)
=>\(\left(\sqrt{x}-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)(vô lý)
Vậy: \(x\in\varnothing\)
cho A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\) và B=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\)
a)rút gọn B
b)tìm x nguyên để C=A.(B-2)CÓ GIÁ TRỊ NGUYÊN
giải hộ e với
ĐKXĐ: \(x\ge0,x\ne4\)
a) \(B=\dfrac{2\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
b) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{2\sqrt{x}+2-2\sqrt{x}-4}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\left(\sqrt{x}-2\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Kết hợp ĐKXĐ:
\(\Rightarrow x\in\left\{0;1;9;16\right\}\)
Cho biểu thức E= \(\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}\):\(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right]\)
a) Rút gọn biểu thức
b) tìm gt của x để E>1
c) tìm giá trị nhỏ nhất của E để E >1
\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)
\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)
Bài 1: Cho A=\(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\div\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\) (x≥0; x≠9)
a, Rút gọn A
b, Tính A khi \(x=7+4\sqrt{3}\)
c, Tìm x để A=\(\dfrac{3}{5}\)
d, Tìm x để A>1
e, Tìm x∈Z để A∈Z
(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x}+3}.\)
(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).
Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)
\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)
(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)
\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).
Vậy: \(x\in\varnothing.\)
(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))
\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)
(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)
Vậy: \(x=0.\)