\(E=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right),\left(a>0,a\ne1\right).\)
1. Rút gon E
2. Tìm \(a\) để \(E=-4\)
\(ChoQ=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)
a, rút gọn
b, chứng minh nếu 0<x<1 thì Q>0
c, tìm GTLN của Q
\(ChoA=\frac{1}{2\left(1+\sqrt{x}+2\right)}+\frac{1}{2\left(1-\sqrt{x}+2\right)}\)
a, tìm x để a có nghĩa
b, rút gon A
c, tìm X nguyên để A nguyên
\(ChoA=\left(\frac{\sqrt{a}}{\sqrt{a-1}}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2}{a-1}\right)\)
a, Rút gọn A
b, tính A Khi a=3+\(2\sqrt{2}\)
C/m biểu thức
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)=1\)(a,b>0,a\(\ne\)0
b)\(\frac{a-b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\left(a,b>0,a\ne b\right)\)
c)\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\left(a>0,a\ne1\right)\)
d)\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)=\left(1-a\right)^2\left(a\ge0,a\ne1\right)\)
Giải giúp mk với. THứ 3 tuần sau là phải nộp rồi
Rút gọn và tính giá trị biểu thức: a, \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
b, \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
c, \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
d,\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)
e,\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\left(x\ne1,y\ne1,y>0\right)\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
\(\(d)\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)\)
\(\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\)\)
\(\(=\frac{|\sqrt{x}-1|}{|\sqrt{x}+1|}\)\)
\(\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)\)( vì \(\(x\ge0\)\))
_Minh ngụy_
Rút gọn:
a) \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1-\sqrt{x}}{1-x}\right)^2\left(x\ge0,x\ne1\right)\)
b) \(B=\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\left(a\ge0,a\ne2,a\ne4\right)\)
c) \(C=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\left(x>0,x\ne1\right)\)
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
\(P=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\frac{1}{2a\sqrt{a}}\left(vớia>0,a\ne1\right)\)
a) Rút gọn biểu thức P.
b) Tìm giá trị của a để P = a.
a)
\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}=\frac{2}{a-1}\)
b) \(\frac{2}{a-1}=a\Rightarrow a^2-a-2=0\)
Ta có: 1+1+(-2)=0, nên pt có 2 nghiệm a1=-1<0 (không thỏa mãn đk)=> loại
a2=2(thỏa mãn đk)=> chọn
Vậy a=2 thì P=a
\(B=\left(\frac{a\sqrt{a}+1}{\sqrt{a}+1}\right):\left(a-1\right)+\frac{2a+\sqrt{a}+1}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}vớia>1\)
\(C=\left(\frac{X-1}{\sqrt{X}-1}+\frac{\sqrt{X^3}-1}{1-X}\right)-\left(\frac{\left(X-1\right)^2+\sqrt{X}}{\sqrt{X}+1}\right)vớiX>0,X\ne1\)
\(D=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}vớix>0,x\ne1\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
Mấy cái này chỉ có nhân lên rồi rút gọn thôi ah. Nên mình cho bạn đáp án để kiểm tra lại thôi ah
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a) Rut gon voi \(a>0,a\ne4;a\ne1\)
b) Tim gia tri cua a de Q duong
a) Q=\(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)=\(\frac{\sqrt{a}-2}{3\sqrt{a}}\) b) Ta thấy \(3\sqrt{a}>0\), để Q dương thì \(\sqrt{a}-2>0\Rightarrow a>4\)
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)....\)
a) Rút gọn A, b) tìm a để A<0.c) Tìm A để A=-2
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\).\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)
=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)
=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)
=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)
=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{-\left(a-1\right)}{1}\)= - a + 1
hok tốt
Bài 1:Rút gọn
\(a,\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(b,\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(c,\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)\times\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\left(a\ne1;a\ge0\right)\)
Bài 2: Rút gọn biểu thức
\(P=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)