KO B DUNG GIQI

giúp với

\(P=\frac{1}{1+a^2+b^2}+\frac{3}{2ab}=\left(\frac{1}{1+a^2+b^2}+\frac{\frac{1}{9}}{2ab}\right)+\frac{26}{18ab}\)

\(\ge\frac{\left(1+\frac{1}{3}\right)^2}{1+a^2+b^2+2ab}+\frac{26}{\frac{9\left(a+b\right)^2}{2}}\ge\frac{\frac{16}{9}}{2}+\frac{26}{\frac{9}{2}}=\frac{20}{3}\)

`Answer:`

Theo giả thiết: `\triangleABC` đồng dạng với `\triangleMNP`

\(\Rightarrow k=\frac{AB}{MN}\) hay \(\frac{2}{5}=\frac{AB}{MN}\)

\(\Rightarrow\frac{MN}{AB}=\frac{5}{2}\)

`->` Chọn C.

Đáp án : C

`Answer:`

\(P=8x^2+3y^2-8xy-6y+21\)

\(=\left(8x^2-8xy+2y^2\right)+y^2-6y+9+12\)

\(=2.\left(4x^2-4xy+y^2\right)+\left(y-3\right)^2+12\)

\(=2.\left(2x-y\right)^2+\left(y-3\right)^2+12\)

\(\Rightarrow P\ge2.0+0+12=12\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x=\frac{3}{2}\\y=3\end{cases}}\)

Vì \(x< 2019\)\(\Leftrightarrow x-2019< 0\)\(\Rightarrow\left|x-2019\right|=2019-x\)

Vậy \(\left|x-2019\right|+x-2018=2019-x+x-2018=1\)với \(x< 2019\)

Với x < 2019 ta có :

| x - 2019 | + x - 2018 = -( x - 2019 ) + x - 2018 = 2019 - x + x - 2018 = 1

`Answer:`

a. \(\frac{2x-10}{4}=5+\frac{2-3x}{6}\)

\(\Leftrightarrow\frac{3\left(2x-10\right)}{12}=\frac{60}{12}+\frac{2\left(2-3x\right)}{12}\)

\(\Leftrightarrow3\left(2x-10\right)=60+2\left(2-3x\right)\)

\(\Leftrightarrow6x-30=60+4-6x\)

\(\Leftrightarrow6x+6x=64+30\)

\(\Leftrightarrow12x=94\)

\(\Leftrightarrow x=\frac{47}{6}\)

b. \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow\frac{1\left(x^2+x+1\right)+\left(2x^2-5\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)

\(\Leftrightarrow x^2+2x^2+x-4x+1-5+4=0\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\text{(Loại)}\end{cases}}\)

c. \(\frac{15x}{x^2+3x-4}-1=12\left(\frac{1}{x+4}+\frac{1}{3x-3}\right)\)

Điều kiện xác định: `x^2+3x-4=(x-1)(x+4)\ne0<=>x\ne1;x\ne-4`

`x+4\ne0<=>x\ne-4`

`3x+3\ne0<=>x\ne-1`

\(\Leftrightarrow\frac{3\left(x+4\right)+12\left(x-1\right)}{\left(x-1\right)\left(x+4\right)}-1=\frac{12}{x+4}+\frac{4}{x+1}\)

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

\(\Leftrightarrow\frac{3}{x-1}-1=\frac{4}{x+1}\)

\(\Leftrightarrow3\left(x+1\right)-\left(x+1\right)\left(x-1\right)=4\left(x-1\right)\)

\(\Leftrightarrow x^2+x=8\)

\(\Leftrightarrow4x^2+4x=32\)

\(\Leftrightarrow\left(2x+1\right)^2=33\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=-\sqrt{33}\\2x+1=\sqrt{33}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1+\sqrt{33}}{2}\\x=\frac{\sqrt{33}-1}{2}\end{cases}}\)