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-( -x + 13 - 142 ) + 18 = 55
-( -x - 129) + 18 = 55
x + 129 + 18 = 55
x + 147 = 55
x = 55 - 147
x = -92
b, 25 - 3.(6 -x ) = 22
3(6-x) = 25 - 22
3(6-x) = 3
6 - x = 1
x =5
c, [ ( 2x - 11 ) :3 +1] .5 = 20
( 2x - 11) : 3 + 1 = 20 : 5
(2x - 11) : 3 + 1 = 4
( 2x - 11) : 3 = 4 - 1
(2x - 11 ) : 3 = 3
2x - 11 = 3 x 3
2x - 11 = 9
2x = 9 + 11
2x = 20
x = 10
d, 3(x+5) - x - 11 = 24
3(x+5) - x = 24 +11
3x + 15 - x = 35
2x = 35 - 15
2x = 20
x = 10
Giải:
a) \(\dfrac{-5}{8}=\dfrac{x}{16}\)
\(\Rightarrow x=\dfrac{16.-5}{8}=-10\)
\(\dfrac{3x}{9}=\dfrac{2}{6}\)
\(\Rightarrow3x=\dfrac{2.9}{6}=3\)
\(\Rightarrow x=1\)
b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
\(\Rightarrow x+3=\dfrac{1.15}{3}=5\)
\(\Rightarrow x=2\)
\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)
\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\)
\(\Rightarrow x=10\)
c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\)
\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\)
\(\Rightarrow x=0\)
\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\)
\(\Rightarrow y=\dfrac{-12.24}{18}=-16\)
\(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\)
\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\)
\(\Rightarrow x=-29\)
\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\)
\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\)
d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\)
\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\)
\(\Rightarrow x\in\left\{-3;-2;-1\right\}\)
\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\)
\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\)
\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\)
\(\Rightarrow x\in\left\{-1;0;1;2\right\}\)
e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\)
\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\)
\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\)
\(\Rightarrow5x+230=100x+40\)
\(\Rightarrow5x-100x=40-230\)
\(\Rightarrow-95x=-190\)
\(\Rightarrow x=-190:-95\)
\(\Rightarrow x=2\)
\(y\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\)
\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\)
\(\Rightarrow y^2+5=86\)
\(\Rightarrow y^2=86-5\)
\(\Rightarrow y^2=81\)
\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\)
Chúc bạn học tốt!
a) \(\dfrac{x}{-6}=\dfrac{-15}{45}\)
\(\dfrac{-x}{6}=\dfrac{-15}{45}\)
\(\dfrac{x}{6}=\dfrac{15}{45}\)
\(x=\dfrac{\left(15\cdot6\right)}{45}\)
\(x=2\)
b) \(\dfrac{x}{5}=\dfrac{16}{25}\)
\(x=\dfrac{\left(16\cdot5\right)}{25}\)
\(x=\dfrac{80}{25}\)
\(x=\dfrac{16}{5}\)
c) \(\dfrac{5}{x-3}=\dfrac{20}{-12}\)
\(x-3=\dfrac{\left(5\cdot-12\right)}{20}\)
\(x-3=-3\)
\(x=\left(-3\right)+3\)
\(x=0\)
d) \(\dfrac{2}{5}\cdot x=\dfrac{6}{35}\)
\(x=\dfrac{6}{35}\div\dfrac{2}{5}\)
\(x=\dfrac{3}{7}\)
-|x|=11
\(\Rightarrow\)|x|=-11
Mà |x|\(\ge\)0
\(\Rightarrow x\in\varnothing\)
Vậy không tìm được giá trị của x thỏa mãn bài toán.
-|x|=-24
|x|=24
\(\Rightarrow\orbr{\begin{cases}x=-24\\x=24\end{cases}}\)
Vậy \(x\in\left\{\pm24\right\}\)
|x|=(-3).(-6)
|x|=18
\(\Rightarrow\orbr{\begin{cases}x=-18\\x=18\end{cases}}\)
Vậy \(x\in\left\{\pm18\right\}\)
|x|=(-5).20
|x|=-100
Mà \(\left|x\right|\ge0\)
\(\Rightarrow x\in\varnothing\)
Vậy không tìm được giá trị của x thỏa mãn bài toán.
-|x|=11
=>|x|=-11
+)Ta có:|x|\(\ge\)0
Mà -11<0
=>Không tìm dc x thỏa mãn
Vậy \(x\in\varnothing\)
Mấy phần khác bn làm tương tự nha
Chúc bn học tốt