B=\(\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}}\right)\) tìm x \(\in N\) để B\(\ge\frac{1}{2}\)
1. A= \(\left(\sqrt{x}-\frac{x+2}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)
a. Rút gọn A
b. Tìm x để A<0
c. Tìm giá trị nhỏ nhất A.
2. M=\(\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{x+4}{x+\sqrt{x}+1}\right)\)
a. Rút gọn M
b. Tìm số nguyên x để M có giá trị nguyên
3. N=\(\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{a.b}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{a.b}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)
a. Rút gọn N
b. Tính N khi a=\(\frac{2}{2-\sqrt{3}}\)
c. Tìm số nguyên a để N có giá trị nguyên
Gíup mình với. Cảm ơn nhiều ạ.
Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
1, A=\(\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{2}{\sqrt{x}+1}\right):\frac{x-1}{\sqrt{x}}\) với x > 0
a, Rút gọn
b, Tìm x nguyên nhỏ nhất để A < 0
c, Tìm \(x\in Z\) để \(A\in Z\)
2, Rút gọn: \(\left(\frac{14}{\sqrt{14}}+\frac{\sqrt{12}+\sqrt{30}}{\sqrt{5}+\sqrt{2}}\right).\sqrt{5-\sqrt{21}}\)
3, Cho \(\left|x\right|< 1,\left|y\right|< 1\). Chứng minh \(\frac{1}{1-x^2}+\frac{1}{1-y^2}\ge\frac{2}{1-xy}\)
Bạn nào giúp mk thứ 2 phải nộp rồi!!!
1, A= \(\frac{\sqrt{x}+4}{\sqrt{x}-1}\) B= \(\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\left(x\ge0,x\ne1\right)\)
Tìm x để \(\frac{A}{B}\ge\frac{x}{4}+5\)biết B= \(\frac{1}{\sqrt{x}-1}\)
2, A= \(\frac{4\left(\sqrt{x}+1\right)}{25-x}\) B= \(\left(\frac{15-5x}{x-25}+\frac{2}{\sqrt{x}+5}\right):\frac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0,x\ne25\right)\)
Tìm giá trị nguyên của x để P= A.B đặt giá trị nguyên lớn nhất
GIÚP MK VỚI! THANKS
Câu 1:
\(\frac{A}{B}\ge\frac{x}{4}+5\Leftrightarrow\frac{\sqrt{x}+4}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\ge\frac{x}{4}+5\)
\(\Rightarrow\sqrt{x}+4\ge\frac{x}{4}+5\Rightarrow x-4\sqrt{x}+4\le0\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2\le0\Rightarrow\sqrt{x}-2=0\Rightarrow x=4\)
Câu 2:
Bạn coi lại đề, biểu thức B không hợp lý
cho bt p=\(\left(\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}}{x-1}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right)\)
a) rg p
b) tìm x để \(\frac{1}{p}-\frac{\sqrt{x}+1}{8}>\)hoặc bằng 1
cho \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right).\)
Tìm x để A \(\ge\frac{3}{2}\)
Ta có: \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\) ( ĐKXĐ: \(x>0,\)\(x\ne0,\)\(x\ne1\))
\(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\left(\frac{2\sqrt{x}}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
Để \(A\ge\frac{3}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{3}{2}\ge0\)
\(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}+3}{2.\left(\sqrt{x}-1\right)}\ge0\)
\(\Leftrightarrow\frac{5-\sqrt{x}}{2.\left(\sqrt{x}-1\right)}\ge0\)
+ TH1: \(\hept{\begin{cases}5-\sqrt{x}\ge0\\2\sqrt{x}-2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\le5\\\sqrt{x}\ge1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\le25\\x\ge1\end{cases}}\)\(\Rightarrow\)\(1\le x\le25\)\(\left(TM\right)\)
+ TH2: \(\hept{\begin{cases}5-\sqrt{x}\le0\\2\sqrt{x}-2\le0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\ge5\\\sqrt{x}\le1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge25\\x\le1\end{cases}}\)\(\left(L\right)\)
\(\Rightarrow\)\(1\le x\le25.\)Kết hợp ĐKXĐ: \(x\ne1\)
\(\Rightarrow\)\(1< x\le25\)
Vậy để \(A\ge\frac{3}{2}\)\(\Leftrightarrow\)\(1< x\le25\)
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
\(B=\left(\frac{\sqrt{x}+2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x-3}}+\frac{3\sqrt{x}-1}{\left(\sqrt{x-1}\right)\left(\sqrt{x}-3\right)}\right):\left(1-\frac{1}{\sqrt{x}-1}\right)\)
a) Rút gọn biểu thức B
b) Tìm x \(\in\) Zđể B nhận giá trị nguyên
1. Chứng minh \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
2. a) Tính \(A=\frac{2b.\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(a,b>0\right) \)
b) Tính \(B=\frac{xy-\sqrt{x^2-1}.\sqrt{y^2-1}}{xy+\sqrt{x^2-1}.\sqrt{y^2-1}}\) với \(x=\frac{1}{2}\left(a+\frac{1}{a}\right);y=\frac{1}{2}\left(b+\frac{1}{b}\right)\left(a,b\ge1\right)\)
3. Cho x,y thỏa mãn \(xy\ge0\). Tính \(B=\left(\left|\sqrt{xy}+\frac{x}{2}+\frac{y}{2}\right|-\left|x\right|\right)+\left(\left|\sqrt{xy}-\frac{x}{2}-\frac{y}{2}\right|-\left|y\right|\right)\)
4. Cho \(\frac{2x+2\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(x+1\right)^2}=\frac{A}{\sqrt{x}-2}+\frac{B\sqrt{x}+C}{x+1}+\frac{D\sqrt{x}+E}{\left(x+1\right)^2}\). Tìm các số A,B,C,D,E để đẳng thức trên là đúng với mọi x