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Ta có : |x-2| và |y-3| >= 0 => |x-2|+|y-3| >=0
=> |x-2|+|y-3| <=0 <=> |x-2|+|y-3|=0
<=> x-2=0 và y-3=0
<=> x=2 và y=3
k mk nha
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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 a ) \(\left|x-11\right|+11-x=0\)
\(\Leftrightarrow\left|x-11\right|=x-11\)
\(\Leftrightarrow\orbr{\begin{cases}x-11=x-11\\x-11=11-x\end{cases}\Leftrightarrow\orbr{\begin{cases}\forall x\\x=11\end{cases}}}\)
p./s tham khảo nha
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1 phần 9 x 27^n = 3^n
=>(3^3)^n = 3^n x 9
=> 3^3n = 3^(n+2)
=>3n = n +2
=>3n - n = 2
=> 2n = 2
=>n = 1
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\(x^2-30=34\)
\(x^2=34+30\)
\(x^2=64=8^2=\left(-8\right)^2\)
Vậy \(x=8^2\) hoặc \(x=\left(-8\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
x2y + xy - x = 6
x2y + xy - x -1 = 5
xy.(x + 1) - (x + 1) = 5
(x = 1).(xy - 1) = 1.5 = (-1).(-5) = 5.1 = (-5).(-1)
Ta có bảng giá trị;
x+1 | 1 | -1 | 5 | -5 |
x | 0 | -2 | 4 | -6 |
xy - 1 | / | -5 | / | -1 |
y | / | 2 | / | 0 |
Vậy (x;y) = (-2;2) ; (-6;0)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(4\left(x-1\right)^{100}-3^{100}=3^{101}\)
\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{101}+3^{100}\)
\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{100}\left(3+1\right)\)
\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{100}.4\)
\(\Leftrightarrow\)\(\left(x-1\right)^{100}=3^{100}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^{100}=3^{100}\\\left(x-1\right)^{100}=\left(-3\right)^{100}\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3+1\\x=-3+1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy \(x=-2\) hoặc \(x=4\)
Chúc bạn học tốt ~