a)

\(\left|x\right|>5\)

\(\left\{{}\begin{matrix}x>5khix>5\\-x>5khix< -5\end{matrix}\right.\)

a) |x| > 5

=> \(\left\{{}\begin{matrix}x>5\\-x>5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>5\left(TM\right)\\x< -5\left(TM\right)\end{matrix}\right.\)

Vậy \(x>5\) và \(x< -5\).

b) |x| < 3

=> \(\left\{{}\begin{matrix}x< 3\\-x< 3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x< 3\left(TM\right)\\x>-3\left(TM\right)\end{matrix}\right.\)

Vậy \(x< 3\) và \(x>-3\).

Mình chỉ làm câu a) và câu b) thôi nhé.

Chúc bạn học tốt!

\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)

\(=\dfrac{4.4^5.6.6^5}{3.3^5.2.2^5}\)

\(=\dfrac{4^6.6^6}{3^6.2^6}\)

\(=\dfrac{2^6.2^6.2^6.3^6}{3^6.2^6}\)

\(=2^{12}=2^{3^4}=8^4=8^x\)

Vậy x = 4

Vậy x = 4

Ta có: (x-5)(x+6)=0

=> x-5=0 hoặc x+6=0

=> x=5 hoặc x=-6

Nhầm nhầm @@

\(\left\{{}\begin{matrix}x+y=5\\y+z=6\\z+t=7\end{matrix}\right.\) \(\Leftrightarrow x+y+z+t=12\)

\(\Leftrightarrow x+2y+z=11\)

\(\Leftrightarrow\left(x+y+z+t\right)-\left(x+y+y+z\right)=1\)

\(\Leftrightarrow x+y+z+t-x-y-y-z=1\)

\(\Leftrightarrow t-y=1\)

\(\Leftrightarrow y+2z+t=13\)

\(\Leftrightarrow\left(y+z+z+t\right)-\left(x+y+z+t\right)=1\)

\(\Leftrightarrow y+z+z+t-x-y-z-t=1\)

\(\Leftrightarrow z-t=1\)

Ta có:

\(\left\{{}\begin{matrix}z+t=7\\z-t=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z+t+z-t=8\Rightarrow2z=8\Rightarrow z=4\\t=4-1=3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=6-4=2\\x=5-2=3\end{matrix}\right.\)

Vậy...

\(\left\{{}\begin{matrix}x+y=5\\y+z=6\\z+t=7\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)-\left(y+z\right)+\left(z+t\right)=5-6+7\)

\(\Rightarrow x+y-y-z+z+t=6\)

\(\Rightarrow x+t=6\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\y+z=6\\z+t=7\\x+t=6\end{matrix}\right.\)

\(\Rightarrow x+y+y+z+z+t+t+x=5+6+7+6\)

\(\Rightarrow2\left(x+y+z+t\right)=18\)

\(\Rightarrow x+y+z+t=9\)

Thay zô tính

a, - Vì \(x\in Z\Rightarrow x-3\in Z\)

- Vì \(y\in Z\Rightarrow2y+5\in Z\)

\(\Rightarrow x-3;2y+5\inƯ\left(34\right)\)

\(\Rightarrow x-3;2y+5\in\left\{\pm1;\pm2;\pm17;\pm34\right\}\)

Ta có bảng sau :

Vậy các cặp số nguyên (x;y) thỏa mãn là : (5;6);(1;-11);(37;-2);(-31;-3)

b, \(xy+3y=6\Leftrightarrow y\left(x+3\right)=6\)

\(\Rightarrow y;x+3\inƯ\left(6\right)\Rightarrow y;x+3\in\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Ta có bảng sau :

Vậy các cặp số nguyên (x;y) thỏa mãn là : (1;3);(2;0);(3;-1);(6;-2);(-1;-9);(-2;-6);(-3;-5);(-6;-4)

x-30%.x=-6/5

x.1-3/10=-6/5

x.(1-3/10)=-6/5

x.7/10=-6/5

x= -6/5: 7/10

x= -12/7

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ