Cho biểu thức M=[\(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)- \(\left(\sqrt{X}+2\right)\)]\(\dfrac{\left(\sqrt{x}-1\right)^2}{2}\) Tìm điều kiện của x để M có nghĩa và rút gọn M Chứng minh M <=\(\dfrac{1}{4}\)
Cho biểu thức P=\(\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x-3}\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
a.Rút gọn P
b.Tính giái trị của P khi x=\(\dfrac{3-2\sqrt{2}}{4}\)
Sửa đề: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{9}{4}\end{matrix}\right.\)
a) Ta có: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\cdot\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2x+2\sqrt{x}+\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b) Ta có: \(x=\dfrac{3-2\sqrt{2}}{4}\)
\(\Leftrightarrow x=\dfrac{2-2\cdot\sqrt{2}\cdot1+1}{4}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\) vào biểu thức \(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\), ta được:
\(P=\left(3\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}-5\right):\left(2\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}+1\right)\)
\(\Leftrightarrow P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(\Leftrightarrow P=\left(\dfrac{3\cdot\left(\sqrt{2}-1\right)}{2}-\dfrac{10}{2}\right):\left(\sqrt{2}-1+1\right)\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-3-10}{2}:\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-13}{2}\cdot\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{6-13\sqrt{2}}{2}\)
Vậy: Khi \(x=\dfrac{3-2\sqrt{2}}{4}\) thì \(P=\dfrac{6-13\sqrt{2}}{2}\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
Rút gọn:
\(M=1-\left[\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right]\cdot\left[\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right]\)
Giải::
ĐK: x khác +- 1
\(M=1-\left[\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)\left(\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{1}{2}\right)\left(\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}\right]\cdot\left[\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}\right]\)
\(=1-\left[\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)}{\left(1-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{1}{2}\right)}{1-\sqrt{x}+x}\cdot\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}\right]\)
\(=1-\left[\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)}{2}+\dfrac{-x\left(1-\sqrt{x}\right)^2}{2\left(1-\sqrt{x}+x\right)}\right]\)
rồi làm sao nữa ak?? Tớ có quy đồng lên, tính sơ sơ rồi nhưng thấy kq không gọn.
Câu b là : tìm các số nguyên x để M cũng là số nguyên . Nên tớ nghĩ kq sẽ gọn.
NHỜ MẤY CAO NHÂN RA TAY GIÚP VỚI NHAK ^^!
Cho biểu thức P = \(\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) ( với x ≥ 0; x ≠ 1 )
a,Rút gọn biểu thức P
b,Tính giá trị của P khi x = \(6-2\sqrt{5}\)
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
Rút gọn các biểu thức sau:
a) \(\left(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)\)
b) \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\) với x>0
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)
7) cho biểu thức: P=\(\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{5}{x+\sqrt{x}-2}\right)\div\left(1+\dfrac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
a) rút gọn P
b) tính P khi \(x=6-2\sqrt{5}\)
c) tính giá trị của x để P= \(\dfrac{1}{\sqrt{x}}\)
giúp mk vs ah mk cần gấp lắm
đk : \(x\ge0,x\ne1\)
\(=>P=\left[\dfrac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]:\left[\dfrac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right]\)
\(P=\left[\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right].\left[\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\right]\)
\(P=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b,\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thay vào P
\(=>P=\dfrac{2\sqrt{\left(\sqrt{5}-1\right)^2}-1}{\sqrt{\left(\sqrt{5}-1\right)^2}+1}=\dfrac{2\sqrt{5}-3}{\sqrt{5}}\)
c,\(=>\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}=>2x-\sqrt{x}=\sqrt{x}+1\)
\(=>2x-2\sqrt{x}-1=0< =>2\left(x-\sqrt{x}-\dfrac{1}{2}\right)=0\)
\(=>x-\sqrt{x}-\dfrac{1}{2}=>\Delta=1-4\left(-\dfrac{1}{2}\right)=3>0=>\left[{}\begin{matrix}x1=\dfrac{1+\sqrt{3}}{2}\\x2=\dfrac{1-\sqrt{3}}{2}\end{matrix}\right.\)
đối chiếu đk loại x2 còn x1 thỏa
Cho x>1 .Hãy rút gọn đa thức sau:
\(M=1-\left[\left(\dfrac{1}{\sqrt{x}-1}-\sqrt{x-1}\right):\left(\dfrac{1}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x-1}}\right):\dfrac{\sqrt{x}+1.\sqrt{x^2-1}}{\left(x-1\right)\sqrt{x+1}-\left(x+1\right).\sqrt{x-1}}\right]\)
rút gọn hoạc tính giá trị các biểu thức sau
1)1+\(\sqrt{\dfrac{\left(x-1\right)^2}{x-1}}\)
2)\(\sqrt{\left(x-2\right)^2}+\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\)
3)\(\sqrt{m}-\sqrt{m-2\sqrt{m}+1}\)
1: \(1+\sqrt{\dfrac{\left(x-1\right)^2}{x-1}}=1+\sqrt{x-1}\)
2: \(A=\sqrt{\left(x-2\right)^2}+\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\)
=\(\left|x-2\right|+\dfrac{x-2}{\left|x-2\right|}\)
TH1: x>2
A=x-2+(x-2)/(x-2)=x-2+1=x-1
TH2: x<2
A=2-x+(x-2)/(2-x)=2-x-1=1-x
3: \(C=\sqrt{m}-\sqrt{m-2\sqrt{m}+1}\)
\(=\sqrt{m}-\sqrt{\left(\sqrt{m}-1\right)^2}\)
\(=\sqrt{m}-\left|\sqrt{m}-1\right|\)
TH1: m>=1
\(C=\sqrt{m}-\sqrt{m}+1=1\)
TH2: 0<=m<1
\(C=\sqrt{m}+\sqrt{m}-1=2\sqrt{m}-1\)
Cho biểu thức P= \(\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
Rút gọn biểu thức trên
\(P=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
Rút gọn biểu thức dạng chữ:
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(x+\sqrt{x}\right)\) với x ≥0, x ≠1
A= \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}}{4-x}\right):\dfrac{\sqrt{x}+1}{x-4}\) với x ≥0, x ≠ 4
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\right):\dfrac{1}{x+6\sqrt{x}+9}\) với x ≥ 0, x ≠ 9
Hộ vs ạ
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
3.
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)
\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)