Cho biểu thức G=\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\)
Chứng minh rằng nếu 0<x<1 thì G nhận giá trị dương
Cho biểu thức A = \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\):\(\dfrac{\sqrt{x}-1}{2}\) (\(x\ge0\); \(x\ne1\)). Chứng minh rằng \(A>0\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
Cho biểu thức \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) và \(P=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\times\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x>0,x\ne1\)
1) Tính giá trị của biểu thức A khi x = 9
2) Chứng minh rằng \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
3) Tìm các giá trị của x để \(2P=2\sqrt{x}+5\)
1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)
2: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
3: 2P=2*căn x+5
=>\(\dfrac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\)
=>\(2x+5\sqrt{x}-2\sqrt{x}-2=0\)
=>\(2x+3\sqrt{x}-4=0\)
=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(2\sqrt{x}-1=0\)
=>x=1/4
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 biểu thức : P= \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2}{x-4}\right).\left(\sqrt{x}-1+\dfrac{\sqrt{x}-4}{\sqrt{x}}\right)\) (với x>0; x\(\ne\)4)
1) Chứng minh rằng P=\(\sqrt{x}\)+3
2) Tìm các giá trị của x sao cho P=x+3
`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`
`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`
`=((x+3\sqrtx-4)/(x-4)).((x-4)/\sqrtx))`
`=(x+3\sqrtx)/\sqrtx`
`=(\sqrtx(\sqrtx+3))/\sqrtx`
`=\sqrtx+3(đpcm)`
`2)P=x+3
`<=>\sqrtx+3=x+3`
`<=>x-\sqrtx=0`
`<=>\sqrtx(\sqrtx-1)=0`
Vì `x>0=>\sqrtx>0`
`=>\sqrtx-1=0<=>x=1(tm)`
Vậy `x=1=>\sqrtx+3=x+3`
`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`
`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`
`=((x+3\sqrtx)/(x-4)).((x-4)/\sqrtx))`
`=(x+3\sqrtx)/\sqrtx`
`=(\sqrtx(\sqrtx+3))/\sqrtx`
`=\sqrtx+3(đpcm)`
`2)P=x+3
`<=>\sqrtx+3=x+3`
`<=>x-\sqrtx=0`
`<=>\sqrtx(\sqrtx-1)=0`
Vì `x>0=>\sqrtx>0`
`=>\sqrtx-1=0<=>x=1(tm)`
Vậy `x=1=>\sqrtx+3=x+3`
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
b, chứng minh: A > 0 với mọi x ≥ 0, x ≠ 1
5.Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\).\(\dfrac{\sqrt{x}+1}{\sqrt{x}}\) với x >0,x ≠ 1
a)Chứng minh rằng Q=\(\dfrac{2}{X-1}\)
b)Tìm x ϵ Z để biểu thức A nhận giá trị nguyên
a) \(Q=\) \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne1\right)\)
\(Q=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(Q=\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{2}{x-1}\) \(\left(đpcm\right)\).
b) Để \(Q\in Z\) <=> \(\dfrac{2}{x-1}\in Z\) <=> \(x-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
x -1 | 1 | -1 | 2 | -2 |
x | 2(TM) | 0(ko TM) | 3(TM) | -1(koTM) |
Vậy để biểu thức Q nhận giá trị nguyên thì \(x\in\left\{2;3\right\}\)
Cho biểu thức sau: \(P=\dfrac{x^2-\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
1, Rút gọn P
2, Giả sử \(x< 1\) chứng minh rằng \(P-\left|P\right|=0\)
3, Tìm giá trị nhỏ nhất của P
Chứng minh biểu thức không phụ thuộc biến:
\(\dfrac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\dfrac{5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\) (x >/0)
Biểu thức nguyên đề thế này:
\(\dfrac{2x}{x+3\sqrt{x}+2}+\dfrac{5\sqrt{x}}{x+4\sqrt{x}+3}+\dfrac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
các đại ca xem... thế nào ạ??.....
Mỹ Duyên, nguyen van tuan
1) Chứng minh đẳng thức $\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right) \cdot \sqrt{3+2 \sqrt{2}}=-4$.
2) Rút gọn biểu thức $A=\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right): \dfrac{2}{x+\sqrt{x}-2}$ với $x>0 ; x \neq 1$.
1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)
=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)
=\(\dfrac{-4}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}+1\right)^2}\)
=\(\dfrac{-4\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
=-4
2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)
=\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\left(\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)
=\(\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)
=\(\dfrac{-\sqrt{x}-2}{\sqrt{x}+1}\)
1. (1−5+√2√2+1)⋅√3+2√2=−4√2+1√(√2+1)2=−4(1−5+22+1)⋅3+22=−42+1(2+1)2=−4.
2. Với x>0;x≠1x>0;x≠1 ta có:
A=(√xx+√x−1√x−1):2x+√x−2A=(xx+x−1x−1):2x+x−2
⇔A=(√x√x(√x+1)−1√x−1):2(√x−1)(√x+2)⇔A=(xx(x+1)−1x−1):2(x−1)(x+2)
⇔A=−2(√x−1)(√x+1)⋅(√x−1)(√x+2)2⇔A=−2(x−1)(x+1)⋅(x−1)(x+2)2
⇔A=−(√x+2)√x+1⇔A=−(x+2)x+1. Vạyy với x>0;x≠1x>0;x≠1, ta có A=−(√x+2)√x+1A=−(x+2)x+1.