\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=3\\21\left(x^5+y^5\right)=31.3\left(x^3+y^3\right)\end{matrix}\right.\)

\(\Rightarrow31\left(x^2+xy+y^2\right)\left(x^3+y^3\right)=21\left(x^5+y^5\right)\)

\(\Rightarrow10x^5+31x^4y+31x^3y^2+31x^2y^3+31xy^4+10y^5=0\)

\(\Rightarrow\left(x+y\right)\left(x+2y\right)\left(2x+y\right)\left(5x^2-2xy+5y^2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y=-x\\y=-\dfrac{1}{2}x\\y=-2x\\y=x=0\end{matrix}\right.\) \(\Rightarrow...\)

\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

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

2.

ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)

\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)

\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)

\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)

\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)

\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

l) (x + 9) . (x² – 25) = 0  

<=> (x + 9) . (x – 5) . (x + 5) = 0   

<=> \(\left[{}\begin{matrix}\text{x + 9 = 0}\\x-5=0\\x+5=0\end{matrix}\right.\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy S = \(\left\{-9,5,-5\right\}\)

      e) |x - 4 |< 7         

<=> \(\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.< =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)

Vậy S = \(\left\{11;-3\right\}\)

I,(x+9).(x^2-25)=0

tương đương:x+9=0

                       x^2-25=0

tương đương : x=-9

                       x=5

e,\(\left|x-4\right|\)=7

tương đương x-4=4

                       x-4=-4

tương đương :x=0

                        x=-8

Bài 1: 

l) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

e) Ta có: |x-4|<7

mà \(\left|x-4\right|\ge0\forall x\)

nên \(\left|x-4\right|\in\left\{0;1;2;3;4;5;6\right\}\)

\(\Leftrightarrow x-4\in\left\{0;1;-1;2;-2;3;-3;4;-4;5;-5;6;-6\right\}\)

hay \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

Vậy: \(x\in\left\{4;5;3;6;2;7;1;8;0;9;-1;10;-2\right\}\)

f) Ta có: \(40< 31+\left|x\right|< 47\)

\(\Leftrightarrow\left|x\right|+31\in\left\{41;42;43;44;45;46\right\}\)

\(\Leftrightarrow\left|x\right|\in\left\{10;11;12;13;14;15\right\}\)

hay \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

Vậy: \(x\in\left\{10;-10;11;-11;12;-12;13;-13;-14;14;15;-15\right\}\)

g) Ta có: \(\left|x+3\right|\le2\)

\(\Leftrightarrow\left|x+3\right|\in\left\{0;1;2\right\}\)

\(\Leftrightarrow x+3\in\left\{0;1;-1;2;-2\right\}\)

hay \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

Vậy: \(x\in\left\{-3;-2;-4;-1;-5\right\}\)

a/ -3x=-60+30

=> -3x=-30

=> x=(-30):(-3)

=> x=10

Ban oi phai x,y∈Z thi moi lam dc

b/x.y=7

Vi x,y∈U(7)=(1,-1,7,-7)

Với x=1 => y=7 y=1=>x=7

x=-1 =>x=-7 y=-1=>x=-7

x=7 => x=1 y=7=>x=7

x=-7 => x=-1 y=-7=>x=-1

c/(x-1)(y+2)=-7

x-1 va y+2∈U(-7)=(1,-1,7,-7)

Bạn giải giống câu b/ nhưng x,y trái đâu nhé

d/2x-3=5x-7

=>(-3)+7=5x-2x

=>4=3x

=>x=4/3

e/(-2)²-8x=20

=>4-8x=20

=>4-20=8x

=>-16=8x

=>x=-2

f/17-(-31)+3x=2x-(-50)

=>17+31+3x=2x+50

=>49+3x=2x+50

=> 3x-2x=50-49

=> x=1

g/2y²-16=34

=>2y²=34+16

=>2y²=50

=>y²=50:2

=>y²=25=5²=(-5)²

=>y=5 va y=-5

a) \(\left(-3\right)x-30=-60\)

\(\Leftrightarrow\left(-3\right)x=-60+30\)

\(\Leftrightarrow\left(-3\right)x=-30\)

\(\Leftrightarrow x=\left(-30\right):\left(-3\right)\)

\(\Leftrightarrow x=10\)

b) \(xy=7\)

\(\Rightarrow xy=1.7=7.1=\left(-1\right).\left(-7\right)=\left(-7\right).\left(-1\right)\)

Ta có bảng sau:

KL: Các cặp số (x; y)...

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

\(\Rightarrow\left(x-1\right)\left(y+2\right)=1.\left(-5\right)=\left(-5\right).1=\left(-1\right).5=5.\left(-1\right)\)

Ta có bảng sau:

KL: Các cặp số (x; y)...

d) \(2x-3=5x-7\)

\(\Leftrightarrow2x-5x=-7+3\)

\(\Leftrightarrow-3x=-4\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

e) \(\left(-2\right)^2-8x=20\)

\(\Leftrightarrow4-8x=20\)

\(\Leftrightarrow8x=4-20\)

\(\Leftrightarrow8x=-16\)

\(\Leftrightarrow x=-2\)

f) \(17-\left(-31\right)+3x=2x-\left(-50\right)\)

\(\Leftrightarrow17+31+3x=2x+50\)

\(\Leftrightarrow48+3x=2x+50\)

\(\Leftrightarrow3x-2x=50-48\)

\(\Leftrightarrow x=2\)

g) \(2y^2-16=34\)

\(\Leftrightarrow2y^2=34+16\)

\(\Leftrightarrow2y^2=50\)

\(\Leftrightarrow y^2=50:2=25\)

\(\Leftrightarrow y^2=5^2=\left(-5\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)

\(12-\left|x+7\right|=12\)

\(\Rightarrow\left|x+7\right|=12-12\)

\(\Rightarrow\left|x+7\right|=0\)

\(\Rightarrow x+7=0\)

\(\Rightarrow x=0-7\)

\(\Rightarrow x=-7\)

Bài 3: 

a: =>|x+7|=0

=>x+7=0

hay x=-7

b: =>|x-3|=4

=>x-3=4 hoặc x-3=-4

=>x=7 hoặc x=-1

c: =>x-5<=0

hay x<=5