Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=\dfrac{x+4\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}-2-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{4\sqrt{x}-1+x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x+4\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\)
b: \(B=\dfrac{x\sqrt{x}+26\sqrt{x}-19}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2x+6\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý
Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)
a: ĐKXĐ: x>=0; x<>1
b: \(B=\dfrac{2\sqrt{x}+2+x-\sqrt{x}}{x-1}\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-1}\)
Khi x=3+2căn2 thì \(B=\dfrac{\sqrt{2}+1}{2+2\sqrt{2}}=\dfrac{1}{2}\)
a/ \(M=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}-\left(\sqrt{x}+2\right)\right].\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(=\dfrac{-2\sqrt{x}}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}=\sqrt{x}-x\)
b/ Chứng minh
\(\sqrt{x}-x\le\dfrac{1}{4}\)
\(\Leftrightarrow4x-4\sqrt{x}+1\ge0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)^2\ge0\) (đúng)
a. ĐKXĐ: \(x\ge-1\)
\(y=\sqrt{x^3+1+2\sqrt{x^3+1}+1}+\sqrt{x^3+1-2\sqrt{x^3+1}+1}\)
\(=\sqrt{\left(\sqrt{x^3+1}+1\right)^2}+\sqrt{\left(\sqrt{x^3+1}-1\right)^2}\)
\(=\left|\sqrt{x^3+1}+1\right|+\left|1-\sqrt{x^3+1}\right|\ge\left|\sqrt{x^3+1}+1+1-\sqrt{x^3+1}\right|=2\)
b.
\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{2\left(x-1\right)}{2\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)
c.
\(y=\dfrac{x-2018+1}{\sqrt{x-2018}}=\sqrt{x-2018}+\dfrac{1}{\sqrt{x-2018}}\ge2\sqrt{\dfrac{\sqrt{x-2018}}{\sqrt{x-2018}}}=2\)
a, \(y=\dfrac{\sqrt{x-2}}{x}=\sqrt{\dfrac{1}{x}-\dfrac{2}{x^2}}\ge0\)
\(min=0\Leftrightarrow\dfrac{1}{x}-\dfrac{2}{x^2}=0\Leftrightarrow x=2\)
b, Áp dụng BĐT Cosi:
\(f\left(x\right)=\dfrac{x}{\sqrt{x-1}}=\dfrac{x-1+1}{\sqrt{x-1}}=\sqrt{x-1}+\dfrac{1}{\sqrt{x-1}}\ge2\)
\(minf\left(x\right)=2\Leftrightarrow x=2\)
1234567890-01234567890-=qưertyuiop[]\';;lkjhfgdsazxcvbnm,./\'l;[]7894561230.+-
Sửa đề: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
a: \(A=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để A<0 thì \(\sqrt{x}-1< 0\)
hay 0<x<1
c: Để A là số nguyên thì \(\sqrt{x}+1-2⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{4;0;9\right\}\)
Sửa đề: x căn x+x
\(P=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{x+1}\right)\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)