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Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
d) \(x-\left(-25+x\right)=13-x.\)
\(\Rightarrow x+25+x=13-x\)
\(\Rightarrow x+x+x=13-25\)
\(\Rightarrow3x=-12\)
\(\Rightarrow x=-4\)
e) \(15-\left(30+x\right)=x-\left(27-\text{| }-8\text{| }\right)\)
\(\Rightarrow15-30-x=x-19\)
\(\Rightarrow-15-x=x-19\)
\(\Rightarrow-15+19=x+x\)
\(\Rightarrow4=2x\)
\(x=2\)
f) \(\left(12x-4^3\right).8^3=4.8^4\)
\(\left(12x-2^6\right).2^9=2^2.2^{12}\)
\(12x-2^6=2^2.2^{12}\div2^9\)
\(12x-64=2^5=32\)
\(12x=96\)
\(x=8\)
g) \(\left[119-\left(3^3-10\right)\right].x=2448\)
\(\left[119-\left(27-10\right)\right].x=2448\)
\(\left[119-17\right].x=2448\)
\(102.x=2448\)
\(x=24\)
h)
[( 10 - x ) .2-51] : 3 - 2 = 3
[( 10 - x ) .2-51] : 3 = 3 + 2
[( 10 - x ) .2-51] : 3 = 5
( 10 - x ) .2 - 51 = 5 . 3
( 10 - x ) . 2 - 51 = 15
( 10 - x ) . 2 = 15 + 51
( 10 - x ) . 2 = 66
10 - x = 66 : 2
10 - x = 33
x = 10 - 33
x = -23
i)
(x-12)-15 = (20-7)-(18+x)
x-12-15 = 20-7-18-x
x-27 = -5-x
=> x-27+5+x = 0
2x-25 =0
2x = 0+25 = 25
k)
6) 8(x – |-7|) – 6(x – 2) = |-8|.6 – 50
7) -7(5 – x) – 2(x – 10) = 15
8) 4(x – 1) – 3(x – 2) = -|-5|
\(\frac{x}{7}=\frac{x+1}{14}\Leftrightarrow14x=7x+7\Leftrightarrow7x=7\Leftrightarrow x=1\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)
\(\Leftrightarrow1\le x\le6\Leftrightarrow x=1;2;3;4;5;6\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
\(\Leftrightarrow\frac{1}{2}-\frac{3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
\(\Leftrightarrow0\le x\le5\Leftrightarrow x=0;1;2;3;4;5\)
\(\frac{x}{7}=\frac{x+1}{14}\)
=> \(\frac{x\cdot2}{7\cdot2}=\frac{x+1}{14}\)
=> \(2x=x+1\)
=> \(2x-x-1=0\)
=> \(1x-1=0\)
=> \(x=1\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)
=> \(1\le x\le6\)
=> \(x=\left\{1;2;3;4;5;6\right\}\)
\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)
=> \(0\le x\le5\)
=> \(x=\left\{0;1;2;3;4;5\right\}\)
yc đề bài em ơi
\(\dfrac{6}{8}=\dfrac{15}{x}\)
\(6x=15.8\)
\(x=20\)