Lời giải:

Đặt \(\frac{1}{x-1}=a; \frac{1}{y-1}=b\) thì HPT trở thành:

\(\left\{\begin{matrix} a-3b=-1\\ 2a+4b=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=\frac{1}{2}\\ b=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x-1}=\frac{1}{2}\\ \frac{1}{y-1}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=y=3\)

Vậy HPT có nghiệm $(x,y)=(3,3)$

1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

a) áp dụng BĐT cô-si ta có:

\(y=\frac{x}{2}+\frac{18}{x}\ge2\sqrt{\frac{x}{2}.\frac{18}{x}}=2\sqrt{9}=6\)

Dấu "=" xảy ra khi:

\(\frac{x}{2}+\frac{18}{x}=6\)

\(\Leftrightarrow\frac{x^2}{2x}+\frac{36}{2x}=\frac{12x}{2x}\)

\(\Rightarrow x^2+36=12x\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x=6\)

tương tự mấy câu tiếp theo

ngu người đê anh em

\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)

\(E=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

ĐK : a >= 0 , a khác 1

\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)

1)

Điều kiện: \(x\geq \frac{-1}{2}\)

Bình phương hai vế:

\(x^2+4=(2x+1)^2=4x^2+4x+1\)

\(\Leftrightarrow 3x^2+4x-3=0\)

\(\Leftrightarrow x=\frac{-2\pm \sqrt{13}}{3}\)

Do \(x\geq -\frac{1}{2}\Rightarrow x=\frac{-2+\sqrt{13}}{3}\) là nghiệm duy nhất của pt.

2)

a) \(x^2+x+12\sqrt{x+1}=36\) (ĐK: \(x\geq -1\) )

\(\Leftrightarrow (x^2+x-12)+12(\sqrt{x+1}-2)=0\)

\(\Leftrightarrow (x-3)(x+4)+\frac{12(x-3)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow (x-3)\left[x+4+\frac{12}{\sqrt{x+1}+2}\right]=0\)

Do \(x\geq -1\Rightarrow x+4+\frac{12}{\sqrt{x+1}+2}\geq 3+\frac{12}{\sqrt{x+1}+2}>0\)

Do đó \(x-3=0\Leftrightarrow x=3\) (thỏa mãn)

Vậy pt có nghiệm x=3

b) Đặt \(\left\{\begin{matrix} \sqrt{x^2+7}=a\\ x+4=b\end{matrix}\right.\)

PT tương đương:

\(x^2+7+4(x+4)-16=(x+4)\sqrt{x^2+7}\)

\(\Leftrightarrow a^2+4b-16=ab\)

\(\Leftrightarrow (a-4)(a+4)-b(a-4)=0\)

\(\Leftrightarrow (a-4)(a+4-b)=0\)

+ Nếu \(a-4=0\Leftrightarrow \sqrt{x^2+7}=4\Leftrightarrow x^2=9\Leftrightarrow x=\pm 3\) (thỏa mãn)

+ Nếu \(a+4-b=0\Leftrightarrow a=b-4\)

\(\Leftrightarrow \sqrt{x^2+7}=x\)

\(\Rightarrow x\geq 0\). Bình phương hai vế thu được: \(x^2+7=x^2\Leftrightarrow 7=0\) (vô lý)

Vậy pt có nghiệm \(x=\pm 3\)

Câu 3:

Ta có \(M=\frac{x^2+2000x+196}{x}\)

\(\Leftrightarrow M=x+2000+\frac{196}{x}\)

Áp dụng BĐT AM-GM ta có: \(x+\frac{196}{x}\geq 2\sqrt{196}=28\)

\(\Rightarrow M=x+\frac{196}{x}+2000\geq 28+2000=2028\)

Vậy M (min) =2028. Dấu bằng xảy ra khi \(\left\{\begin{matrix} x=\frac{196}{x}\\ x>0\end{matrix}\right.\Rightarrow x=14\)