Nhầm

\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(\Rightarrow3^2A=3^2+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(\Rightarrow9A-A=\left(9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)

\(\Rightarrow8A=9-\frac{1}{3^{100}}\)

=> n = 100

\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(\Rightarrow3^2A=3^2+\frac{1}{3^4}+\frac{1}{3^6}+...+\frac{1}{3^{102}}\)

\(\Rightarrow9A-A=\left(9+\frac{1}{3^4}+\frac{1}{3^6}+...+\frac{1}{3^{102}}\right)-\left(1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)\)

\(\Rightarrow8A=9-\frac{1}{3^{100}}\)

\(\Rightarrow n=100\)

\(xy+x+2y=5\)

\(\Leftrightarrow x\left(y+1\right)+2y+2=5+2\)

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

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

Ta có bảng:

Vậy các cặp (x;y) là (-1;6) ; (-3;-8) ; (5;0) ; (-9;-2)

Câu 1:

a. Ta có:\(\left|x-2017\right|\ge0\)

\(\Rightarrow2016+\left|x-2017\right|\ge2016\)

\(\Rightarrow A\ge2016\)

Dấu " = " xảy ra <=> x - 2017 = 0 <=> x = 2017

Vậy Min_A = 2016 khi x = 2017

b. Ta có: \(\left(x-10\right)^2\ge0\)

\(\Rightarrow2017+\left(x-10\right)^2\ge2017\)

\(\Rightarrow B\ge2017\)

Dấu " = " xảy ra <=> (x - 10)² = 0 <=> x = 10

Vậy Min_B = 2017 khi x = 10

c.Ta có: \(\left(x-1\right)^4\ge0\)

\(\Rightarrow-50+\left(x-1\right)^4\ge-50\)

\(\Rightarrow C\ge-50\)

Dấu " = " xảy ra <=> (x - 1)⁴ = 0 <=> x = 1

Vậy Min_C = -50 khi x = 1

Câu 2:

a. Ta có: \(-\left(x-10\right)^2\le0\)

\(\Rightarrow-100-\left(x-10\right)^2\le-100\)

\(\Rightarrow M\ge-100\)

Dấu " = " xảy ra <=> ( x - 10)² = 0 <=> x = 10

Vậy Max_M = -100 khi x = 10

b. Ta có: \(-\left(x-5\right)^2\le0\)

\(\Rightarrow2017-\left(x-5\right)^2\ge2017\)

\(\Rightarrow N\ge2017\)

Dấu " = " xảy ra <=> (x - 5)² = 0 <=> x = 5

Vậy Max_N = 2017 khi x = 5

c. Ta có: \(-\left(x+7\right)^4\le0\)

\(\Rightarrow-25-\left(x+7\right)^4\le-25\)

\(\Rightarrow Q\le-25\)

Dấu " = " xảy ra <=> (x + 7)⁴ = 0 <=> x = -7

Vậy Max_Q = -25 khi x = -7

a. a.b = 4

Vậy các cặp (a;b) là (1;4) ; (-1;-4) ; (2;2) ; (-2;-2) ; (4;1) ; (-4;-1)

b. a.b = -10

Vậy các cặp (a;b) là (1;-10) ; (-1;10) ; (2;-5) ; (5;-2) ; (10;-1) ; (-10;1)

4 người 

**** cho tớ nhé Biêtdongsaigon

Đặt \(A=\left|x-8\right|+2016\)

Ta có : \(\left|x-8\right|\ge0\)

\(\Rightarrow\left|x-8\right|+2016\ge2016\)

\(\Rightarrow A\ge2016\)

Dấu " = " xảy ra <=> x - 8 = 0 <=> x = 8

Vậy Min_A = 2016 khi x = 8

\(xy+2x+3y=-6\)

\(\Leftrightarrow x\left(y+2\right)+3y+6=-6+6\)

\(\Leftrightarrow x\left(y+2\right)+3\left(y+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(y+2\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x+3=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

\(\left|x+9\right|.2=10\)

\(\left|x+9\right|=10:2\)

\(\left|x+9\right|=5\)

\(\Rightarrow\left\{\begin{matrix}x+9=5\\-\left(x+9\right)=5\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-4\\-x-9=5\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=-4\\-x=14\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=-4\\x=-14\end{matrix}\right.\)

\(6n+8⋮3n-1\)

\(\Rightarrow6n+8-2\left(3n-1\right)⋮3n-1\)

\(\Rightarrow6n+8-6n+2⋮3n-1\)

\(\Rightarrow10⋮3n-1\)

\(\Rightarrow3n-1\inƯ\left(10\right)\)

\(\Rightarrow3n-1\in\left\{1;2;5;10\right\}\)

Vậy n \(\in\) {1;2}