tìm x, bt
2/x-2/+13=15
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Đặt \(A=x^2+4xy+2y^2-22y+173\)
\(A=\left(x^2+2xy+y^2\right)+\left(y^2-22y+121\right)+52\)
\(A=\left(x+y\right)^2+\left(y-11\right)^2+52\)
\(\left(x+y\right)^2\ge0;\left(y-11\right)^2\ge0\) với mọi x;y => \(A=\left(x+y\right)^2+\left(y-11\right)^2+52\ge52\)
=>minA=52 <=> \(\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-11\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=0\\y-11=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-11\\y=11\end{cases}}\)
Vậy min=52 khi x=-11 và y=11
(2x - 1 ) . (y + 2) = 15
=> 2x - 1 và y + 2 \(\in\)Ư(15)
Ư(15) là 1; 3; 5; 15
Ta có bảng sau:
2x - 1 | 1 | 3 | 5 | 15 |
y + 2 | 15 | 5 | 3 | 1 |
x | 1 | 2 | 3 | 8 |
y | 13 | 3 | 1 | // |
Kiểm tra | Thỏa mãn | TM | TM | Loại |
Vậy các cặp (x;y) TM là: (1;13); (2;3); (3;1)
-12 + x = 5x - 20
=> -12 + 20 = 5x - x
=> 8 = 4x
=> x = 8 : 4
=> x = 2
b) 4x - 10 = 15 - x
=> 4x + x = 15 + 10
=> 5x = 25
=> x = 25 : 5
=> x = 5
a) -12+x=5x-20
5x-x=20-12
4x=8
x=8:4
x=2
Vậy x=2
b) 4x-10=15-x
4x+x=15+10
5x=25
x=25:5
x=5
Vậy x=5
\(a,\Rightarrow\left[{}\begin{matrix}x-1=2x\\1-x=2x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Rightarrow\left[{}\begin{matrix}x+x-2=2\left(x\ge2\right)\\x+2-x=2\left(0\le x< 2\right)\\-x+2-x=2\left(x< 0\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\left(x\ge2\right)\left(tm\right)\\x=0\left(0\le x< 2\right)\left(tm\right)\\x=0\left(x< 0\right)\left(ktm\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
a: Ta có: \(\left|x-1\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x\left(x\ge1\right)\\x-1=-2x\left(x< 1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\dfrac{1}{3}\left(nhận\right)\end{matrix}\right.\)
hộ mik nha
a)\(2.x-\frac{3}{2}=\frac{1}{2}\)
\(2x=\frac{4}{2}\)
\(x=\frac{4}{2}:2\)
\(x=1\)
Vậy x=1
b)\(\frac{8}{13}.x+\frac{5}{13}.x=\frac{1}{10}\)
\(x.\left(\frac{8}{13}+\frac{5}{13}\right)=\frac{1}{10}\)\
\(x.1=\frac{1}{10}\)
\(x=\frac{1}{10}\)
Vậy x=\(\frac{1}{10}\)
c)\(\frac{2}{15}+\frac{7}{15}.x=\frac{1}{15}\)
\(\frac{7}{15}x=\frac{-1}{15}\)
x=\(\frac{-1}{7}\)
Vậy \(x=\frac{-1}{7}\)
\(\frac{3}{2}\cdot x-\frac{3}{2}=\frac{1}{2}\) \(\frac{8}{13}\cdot x+\frac{5}{13}\cdot x=\frac{1}{10}\)
\(\frac{3}{2}\cdot x=\frac{3}{2}+\frac{1}{2}=2\) \(\left[\frac{8}{13}+\frac{5}{13}\right]\cdot x=\frac{1}{10}\)
\(x=2:\frac{3}{2}\) \(1\cdot x=\frac{1}{10}\)
x=\(\frac{4}{3}\) \(x=\frac{1}{10}:1\)
Vậy x=\(\frac{4}{3}\) \(x=\frac{1}{10}\)
Vậy x=\(\frac{1}{10}\)
c,\(\frac{2}{5}+\frac{7}{15}\cdot x=\frac{1}{15}\)
\(\frac{7}{15}\cdot x=\frac{1}{15}-\frac{2}{5}=\frac{-1}{3}\)
\(x=\frac{-1}{3}:\frac{7}{15}\)
\(x=\frac{-5}{7}\)
Vậy x=\(\frac{-5}{7}\)
bạn viết sai đề bài nhé
(x+2)/11+(x+2)/12+(x+2)/13=(x+2)/14+(x+2)15
<=> (x+2)/11+(x+2)/12+(x+2)/13 - (x+2)/14 - (x+2)/15 = 0
<=> (x+2)(1/11+1/12+1/13 - 1/14 - 1/15 ) = 0
vì: (1/11+1/12+1/13 - 1/14 - 1/15 ) khác 0 nên x-2 = 0 => x=2
\(\left(\frac{2}{11\times13}+\frac{2}{13\times15}+\frac{2}{15\times17}+\frac{2}{17\times19}+\frac{2}{19\times21}\right)\times462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)\times462-x=19\)
\(\left(\frac{1}{11}-\frac{1}{21}\right)\times462-x=19\)
\(\frac{10}{231}\times462-x=19\)
\(20-x=19\)
\(\Rightarrow x=20-19=1\)
\(2.\left|x-2\right|+13=15\)
\(\Rightarrow2.\left|x-2\right|=2\)
\(\Rightarrow\left|x-2\right|=1\)
Nếu \(x-2\ge0\) thì \(x\ge2\)
\(\Rightarrow x-2=1\Rightarrow x=3\) (nhận)
Nếu \(x-2< 0\) thì x < 2
\(\Rightarrow-x+2=1\Rightarrow-x=-1\Rightarrow x=1\) (nhận)
Vậy x = {3;1}
2/x-2=15-13=2
2x=2+2=4
x=4/2=2
vậy x = 2
chúc bọn hok tốt