\(D=x^2-4x+4+2015=\left(x-2\right)^2+2015\ge2015\)

Dấu = xảy ra khi x-2=0

=> x=2

Vậy Min D=2015 <=> x=2

vì x=2018 ...hihi

\(\frac{\left(3x-1\right)\left(4x+3\right)}{2}=\left(1-3x\right)\left(2x-5\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(4x+3\right)=\left(1-3x\right)\left(2x-5\right).2\)

\(\Leftrightarrow12x^2+5x-3=12x^2+3x-10\)

\(\Leftrightarrow12x^2+5x+7=-12x^2+34x\)

\(\Leftrightarrow12x^2+5x+7=-12x^2+34x-34x\)

\(\Leftrightarrow12x^2-29x+7=-12x\)

\(\Leftrightarrow12x^2-29x+7=-12x+12x\)

\(\Leftrightarrow24x^2-29x+7=0\)

\(\Rightarrow\hept{\begin{cases}x=\frac{7}{8}\\x=\frac{1}{3}\end{cases}}\)

Trình độ hơi thấp, có gì sai sót mong bạn bỏ qua

\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

<=>   \(\left(3x-2\right)\left(x+1\right)^2.3^2.\left(3x+8\right)+144=0\)

<=>  \(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)   (*)

Đặt  \(3x+3=t\) Khi đó pt (*) trở thành: 

   \(\left(t-5\right)t^2\left(t+5\right)+144=0\)

<=>  \(t^4-25t^2+144=0\)

<=>  \(\left(t-4\right)\left(t-3\right)\left(t+3\right)\left(t+4\right)=0\)

đến đây bn tự giải tiếp nhé

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=>  \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

<=>  \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=>  \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Nhận thấy:   \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)

=>  \(x-105=0\)

<=>  \(x=105\)

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Leftrightarrow x=105\)

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

\(\Leftrightarrow1-x=1-2x\)

\(\Leftrightarrow-x+2x=1-1\)

\(\Leftrightarrow x=0\)

Tương tự ta cũng có \(y=0\)

Khi đó : \(x^2+y^2-xy=0^2+0^2-0\cdot0=0=0^2\left(đpcm\right)\)

Sai đề ạ:

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