Lời giải:

Ta có: \(xy+yz+xz=1\)

\(\Rightarrow \left\{\begin{matrix} x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\\ y^2+1=y^2+xy+yz+xz=(y+z)(y+x)\\ z^2+1=z^2+xy+yz+xz=(z+x)(z+y)\end{matrix}\right.\)

Do đó:

\(\sqrt{\frac{(y^2+1)(z^2+1)}{x^2+1}}=\sqrt{\frac{(y+z)(y+x)(z+x)(z+y)}{(x+y)(x+z)}}=\sqrt{(y+z)^2}=y+z\)

\(\Rightarrow x\sqrt{\frac{(y^2+1)(z^2+1)}{x^2+1}}=x(y+z)\)

Hoàn toàn tt:

\(y\sqrt{\frac{(z^2+1)(x^2+1)}{y^2+1}}=y(x+z)\); \(z\sqrt{\frac{(x^2+1)(y^2+1)}{z^2+1}}=z(x+y)\)

Do đó:

\(A=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)

2.

Áp dụng bất đẳng thức Cauchy - schwarz ( hay còn gọi là bất đẳng thức Cosi ):

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}=\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x = y = z = 1

1: 

Áp dụng bất đẳng thức Cô si:

\(x\left(y+\frac{x}{1+y}\right)+y\left(z+\frac{y}{1+z}\right)+z\left(x+\frac{z}{1+x}\right)\)

\(=\left(x+y+z\right)\left[\left(y+\frac{x}{1+y}\right)+\left(z+\frac{y}{1+z}\right)+\left(x+\frac{z}{1+x}\right)\right]\)

\(=1\left[\left(x+y+z\right)+\left(\frac{x}{1+y}+\frac{y}{1+z}+\frac{z}{1+x}\right)\right]\)

\(=1\left[1+\left(\frac{x+y+z}{1+y+1+z+1+x}\right)\right]\)

\(=1\left[1+\left(\frac{1}{3+\left(x+y+z\right)}\right)\right]\)

\(=1\left[1+\frac{1}{4}\right]\)

\(=1+\frac{5}{4}=\frac{9}{4}\)

Dấu "=" xảy ra khi x = y = z = \(\frac{1}{3}\)

2. áp dạng bất đẳng thức cauchy - schwarz dạng engel

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{3^2}{3+3}=\frac{9}{6}=\frac{3}{2}\)

dấu bằng xay ra khi x=y=z=1

lm bất đẳng thức cô si nhé!!! Thanks

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)

Lời giải:

Ta có

\(xy+yz+xz=1\Rightarrow x^2+1=x^2+xy+yz+xz=(x+y)(x+z)\)

Tương tự: \(\left\{\begin{matrix} y^2+1=(y+z)(y+x)\\ z^2+1=(z+x)(z+y)\end{matrix}\right.\)

Do đó \(A=x\sqrt{\frac{(y+z)(y+x)(x+z)(z+y)}{(x+y)(x+z)}}+y\sqrt{\frac{(z+x)(z+y)(x+y)(x+z)}{(y+z)(y+x)}}+z\sqrt{\frac{(x+y)(x+z)(y+x)(y+z)}{(z+x)(z+y)}}\)

\(\Leftrightarrow A=x(y+z)+y(x+z)+z(x+y)=2(xy+yz+xz)=2\)

Vậy \(A=2\)

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3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)

\(\Leftrightarrow2x-6-3+6x=4+4-4x\)

\(\Leftrightarrow8x-9=8-4x\)

\(\Leftrightarrow8x+4x=8+9\)

\(\Leftrightarrow12x=17\)

\(\Leftrightarrow x=\dfrac{17}{12}\)

Vậy \(x=\dfrac{17}{12}\)

4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)

\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)

\(\Leftrightarrow6x-12-4-4x=12-9x-12\)

\(\Leftrightarrow6x-4-4x=12-9x\)

\(\Leftrightarrow2x-4=12-9x\)

\(\Leftrightarrow2x+9x=12+4\)

\(\Leftrightarrow11x=16\)

\(\Leftrightarrow x=\dfrac{16}{11}\)

Vậy \(x=\dfrac{16}{11}\)

thứ lỗi cho mk , mk không biết làm ; bài này khó quá

chuẩn k chỉnh

mik không bieetsn hà banj

Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)

\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)

\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)