bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 =>  x-1/3=y-2/4=z-3/5 

áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1

do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự

Bài 1: 

a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )

cũng dễ thôi

cảm ơn mọi người nhìu nha!!!

B1: Đk: 5x ≥ 0 => x ≥ 0

Vì |x + 1| ≥ 0 => |x + 1| = x + 1

     |x + 2| ≥ 0 => |x + 2| = x + 2

     |x + 3| ≥ 0 => |x + 3| = x + 3

     |x + 4| ≥ 0 => |x + 4| = x + 4

=> |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x

 => x + 1 + x + 2 + x + 3 + x + 4 = 5x

=> 4x + 10 = 5x

=> x = 10

B2: Ta có: |x - 2018| = |2018 - x|

=> A=|x + 2000| + |2018 - x| ≥ |x + 2000 + 2018 - x| = |4018| = 4018

Dấu " = " xảy ra <=> (x + 2000)(x - 2018) ≥ 0

Th1: \(\hept{\begin{cases}x+2000\ge0\\x-2018\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge-2018\\x\le2018\end{cases}}\Rightarrow-2018\le x\le2018\)

Th2: \(\hept{\begin{cases}x+2000\le0\\x-2018\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\le-2018\\x\ge2018\end{cases}}\)(vô lý)

Vậy GTNN của A = 4018 khi -2018 ≤ x ≤ 2018

B3:

a, Vì |x + 1| ≥ 0 ; |2y - 4| ≥ 0

=> |x + 1| + |2y - 4| ≥ 0

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+1=0\\2y-4=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}\)

Vậy...

b, Vì |x - y + 1| ≥ 0 ; (y - 3)² ≥ 0

 => |x - y + 1| + (y - 3)² ≥ 0 

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y+1=0\\y-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-y=-1\\y=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3=-1\\y=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=3\end{cases}}\)

Vậy...

c, Vì |x + y| ≥ 0 ; |x - z| ≥ 0  ; |2x - 1| ≥ 0 

=> |x + y| + |x - z| + |2x - 1| ≥ 0 

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+y=0\\x-z=0\\2x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=0\\x=z\\x=\frac{1}{2}\end{cases}\Leftrightarrow}}\hept{\begin{cases}\frac{1}{2}+y=0\\x=z=\frac{1}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{-1}{2}\\x=z=\frac{1}{2}\end{cases}}\)

coi lại mới thấy trình bày ngờ-u :)) 

B1: Đk: 5x ≥ 0 => x ≥ 0

=> x + 1 > 0 => |x + 1| = x + 1

=> x + 2 > 0 => |x + 2| = x + 2 

=> x + 3 > 0 => |x + 3| = x + 3 

=> x + 4 > 0 => |x + 4| = x + 4 

Ta có:  |x + 1| + |x + 2| + |x + 3| + |x + 4| = 5x

=> .... Làm tiếp như dưới

\(B=\left(x^2+1\right)\left(y^2+1\right)-\left(x-4\right)\left(x+4\right)-\left(y-5\right)\left(y+5\right)\\ B=x^2y^2+x^2+y^2+1-x^2+16-y^2+25\\ B=x^2y^2+41\ge41\)

Dấu "=" xảy ra khi \(x^2y^2\Leftrightarrow x=y=0\)

Vậy \(MaxB=41\Leftrightarrow x=y=0\)

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\\ A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\\ A=\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi

\(\left(x^2+5x\right)^2=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(MaxA=-36\Leftrightarrow x\in\left\{0;-5\right\}\)

Bài 1 :

\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)

\(\Leftrightarrow x=-23\)

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)

Ta co: \(\hept{\begin{cases}x^2-y+\frac{1}{4}=0\\y^2-x+\frac{1}{4}=0\end{cases}}\)

\(\Rightarrow x^2-x+\frac{1}{4}+y^2-y+\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow x=y=\frac{1}{2}}\)

Vậy \(x=y=\frac{1}{2}\)

Ta có: \(\hept{\begin{cases}x^2-y+\frac{1}{4}=0\\y^2-x+\frac{1}{4}=0\end{cases}}\)

\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow x=y=\frac{1}{2}}\)

Vậy \(x=y=\frac{1}{2}\)

2: Tọa độ giao điểm là:

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

Ta có \(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Leftrightarrow xy\ge1\)

\(A=\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{x^2+y^4+2x^2y}\\ \le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}=\dfrac{1}{4xy}+\dfrac{1}{4xy}\\ \le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

Dấu \("="\Leftrightarrow x=y=1\)