Cho biểu thức:
A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}}{x-1}\)
a.Rút gọn biểu thức A
b.Tìm m để phương trình \(mA=\sqrt{x}-2\) có 2 nghiệm phân biệt
Cho biểu thức:
A=\(\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)(với x>0;x\(\ne\)4)
a.Rút gọn A
b.Tìm x để A<-1
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\left(\dfrac{2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(=\dfrac{-4}{\sqrt{x}+2}\)
Lời giải:
a)
\(A=\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right].\frac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\frac{\sqrt{x}+2+2\sqrt{x}+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{4\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{2-\sqrt{x}}{\sqrt{x}}=\frac{-4}{\sqrt{x}+2}\)
b)
$A< -1\Leftrightarrow \frac{-4}{\sqrt{x}+2}+1< 0$
$\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+2}< 0$
$\Leftrightarrow \sqrt{x}-2< 0\Leftrightarrow 0\leq x< 4$
Kết hợp với ĐKXĐ suy ra $0< x< 4$
ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+2+2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{-2}{\sqrt{x}+2}\)
b/ \(A< -1\)
\(\Leftrightarrow\dfrac{-2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\)
Vậy..
Cho biểu thức A=(\(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\)):\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)với x\(\ge\)0 x\(\ne\)1
a.Rút gọn biểu thức A
b.Tìm giá trị của x để A=\(\dfrac{1}{3}\)
`A=1/3`
`<=>3\sqrtx-3=\sqrtx`
`<=>2\sqrtx=3`
`<=>x=9/4`
c1: Rút gọn biểu thức A=\(\left(\dfrac{1}{x-2\sqrt{x}}-\dfrac{2}{6-3\sqrt{x}}\right):\left(\dfrac{2}{3}+\dfrac{1}{\sqrt{x}}\right)\)
c2: Cho phương trình: \(x^2-2\left(2m-1\right)x+m^2-4m=0\left(1\right)\)
Tìm m để phương trình (1) có hai nghiệm phân biệt x1, x2 thoả mãn hệ thức \(x_1+x_2=\dfrac{-8}{x_1+x_2}\)
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
cho biểu thức
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a.rút gọn Q
b.tìm số nguyên x để Q có gtri nguyên
\(\left(đk:x\ne\pm1\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}}{x-1}\)
Cho A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{1-x}\) với x≥0,x≠1
a) Rút gọn A
b) Tìm m để phương trình mA=\(\sqrt{x}-2\) có 2 nghiệm phân biệt
c) Tìm x để A nhận giá trị nguyên
cho biểu thức
A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a.rút gọn biểu thức A
b. chứng minh 0<A<2
Điều kiện: \(x\ge0,x\ne1\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Ta có \(x+\sqrt{x}+1=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\Rightarrow A>0\)
Lại có: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}+1>0;x+\sqrt{x}>0\) với mọi \(x\in TXĐ\)
\(\Rightarrow A-2< 0\Rightarrow A< 2\)
Vậy \(0< A< 2\)
Cho hai biểu thức A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)và B=(\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1})_{ }\div\dfrac{\sqrt{x}}{\sqrt{x}-1}\) (X>0,X≠1)
a.rút gọn biểu thức B
B.Tìm x để giá trị của A.B<0
`a,`
\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\\ =\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{4}{\sqrt{x}+1}\)
`b,` Để `A *B<0` ta có :
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{4}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.4>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)
Kết hợp với đkxđ ta có : \(0< x< 1\)
cho biểu thức
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
a.rút gọn Q
b.tìm số nguyên x để Q có gtri nguyên
đáp an: a.Q=\(\dfrac{2}{x-1}\)
b.x=-1;x=0;x=2;x=3 thì Q\(\in Z\)
mk chỉ bt đáp án chứ ko biết cách giải
ĐKXĐ: \(x>0;x\ne1\)
\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{1}{\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right).\dfrac{1}{\sqrt{x}}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{1}{\sqrt{x}}=\dfrac{2}{x-1}\)
b.
Để \(Q\in Z\Rightarrow2⋮\left(x-1\right)\Rightarrow x-1=Ư\left(2\right)\)
\(\Rightarrow x-1=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x=\left\{-1;0;2;3\right\}\)
Kết hợp ĐKXĐ: \(\Rightarrow x=\left\{2;3\right\}\)
(Đáp án của đề bài đã quên mất ĐKXĐ ban đầu nên ko loại 2 giá trị \(x=-1;x=0\))
1.cho biểu thức A=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\) với x>0,x\(\ne4\)
a.rút gọn biểu thức M
b.tính giá trị của M khi x=3+2\(\sqrt{2}\)
c.tìm giá trị của x để M>0
a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b) Thay x=3+2\(\sqrt{2}\)
A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)
A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0
\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)