x vô ngiệm

qua học 24 mà coi

\(3a^2+4ab+b^2=3a^2+3ab+ab+b^2=3a\left(a+b\right)+b\left(a+b\right)=\left(3a+b\right)\left(a+b\right)\)

xong AM -GM

*Sửa đề: tìm  GTNN

\(A=\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\)

\(=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)

Áp dụng BĐT AM-GM ta có:

\(\frac{\sqrt{c-2}}{c}=\frac{\sqrt{2\left(c-2\right)}}{\sqrt{2}c}\ge\frac{\frac{2+c-2}{2}}{\sqrt{2}c}=\frac{\frac{c}{2}}{\sqrt{2}c}=\frac{1}{2\sqrt{2}}\)

TƯơng tự cho 2 BĐT còn lại ta cũng có:

\(\frac{\sqrt{a-3}}{a}\ge\frac{1}{2\sqrt{3}};\frac{\sqrt{b-4}}{b}\ge\frac{1}{2\sqrt{4}}\)

Suy ra \(A\ge\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}\right)\)

a) \(A=\left[\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{1}{x+1}+\frac{x}{x-1}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)

\(A=\left[\frac{\left(x+1-x+1\right)\left(x-1+x-1\right)}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)

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

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

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

\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left(\frac{x+1}{x-1}\right)\)

\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{x+1}\)

\(A=\frac{4x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x+1\right)}\)

\(A=\frac{4x}{2\left(x+1\right)}\)

\(A=\frac{2x}{x+1}\)

b) Thay A = -3 vào biểu thức a ta được:

\(\frac{2x}{x+1}=-3\)

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

\(\Rightarrow\)\(2x=-3x-3\)

\(\Rightarrow\)\(2x+3x=-3\)

\(\Rightarrow\)\(5x=-3\)

\(\Rightarrow\)\(x=-\frac{3}{5}\)

Vậy khi \(x=-\frac{3}{5}\)thì biểu thức A có giá trị là -3

\(\sqrt{x^2}-2x=5\)

\(x-2x=5\)

\(x\left(1-2\right)=5\)

\(\left(-1\right)\cdot x=5\)

\(x=5\div\left(-1\right)\)

\(x=-5\)

a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)

       = \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)

       = \(\frac{-2\sqrt{6}}{2}\)

       = \(-\sqrt{6}\)