140 : ( 5x - 15 ) = 4
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Ta có: \(xy=140\Rightarrow x=\dfrac{140}{y}\)
\(5x-y=15\)
\(\Leftrightarrow5.\dfrac{140}{y}-y=15\)
\(\Leftrightarrow\dfrac{700}{y}-\dfrac{y^2}{y}=15\)
\(\Leftrightarrow\dfrac{700-y^2}{y}=15\)
\(\Rightarrow y=20\)
\(\Rightarrow x=\dfrac{140}{y}=\dfrac{140}{20}=7\)
ta có: 5x - y =15 => y = 5x - 15
=> x.(5x - 15) = 140
=> 5x^2 - 15x -140 = 0
=> 5.(x^2 -3x -28) = 0
=> x^2 - 4x + 7x -28 = 0
=> x(x-4) +7.(x-4) = 0
=> (x+7).(x-4)=0
=> x= -7 hoặc x=4
Nếu x=-7 => y= -50
Nếu x=4 => y=5
vậy hệ phương trình có cặp nghiệm là: (x;y)=(-7;-50) hoặc( x;y) = ( 4;5)
Bài 1:
a) Ta có: \(82-7\left(3x-4\right)=47\)
\(\Leftrightarrow82-21x+28-47=0\)
\(\Leftrightarrow-21x+63=0\)
\(\Leftrightarrow-21x=-63\)
hay x=3(nhận)
Vậy: x=3
b) Ta có: \(97+4\left(5x-7\right)=129\)
\(\Leftrightarrow97+20x-28-129=0\)
\(\Leftrightarrow20x-60=0\)
\(\Leftrightarrow20x=60\)
hay x=3(nhận)
Vậy: x=3
c) Ta có: \(\left(7x-13\right)\cdot27-12=15\)
\(\Leftrightarrow189x-351-12-15=0\)
\(\Leftrightarrow189x-378=0\)
\(\Leftrightarrow189x=378\)
hay x=2(nhận)
Vậy: x=2
d) Ta có: \(\left(2x+3\right)\cdot13+23=140\)
\(\Leftrightarrow26x+39+23-140=0\)
\(\Leftrightarrow26x-78=0\)
\(\Leftrightarrow26x=78\)
hay x=3(nhận)
Vậy: x=3
đ) Ta có: \(52x+8x-5x=70\)
\(\Leftrightarrow55x=70\)
\(\Leftrightarrow x=\frac{70}{55}\)(loại)
Vậy: x∈∅
e) Ta có: \(19x-3x-x=60\)
\(\Leftrightarrow15x=60\)
hay x=4(nhận)
Vậy: x=4
g) Ta có: \(7\left(3x+1\right)-5\left(3x+1\right)=74\)
\(\Leftrightarrow2\left(3x+1\right)=74\)
\(\Leftrightarrow3x+1=37\)
\(\Leftrightarrow3x=36\)
hay x=12(nhận)
Vậy: x=12
h) Ta có: \(5\left(3x-1\right)+7\left(3x-1\right)=96\)
\(\Leftrightarrow12\left(3x-1\right)=96\)
\(\Leftrightarrow3x-1=8\)
\(\Leftrightarrow3x=9\)
hay x=3(nhận)
Vậy: x=3
120 - { 300: [ 40 - (2.562 - 262.5)] + 10}
= 120 - { 300 : [ 40 - (-186)] + 10}
= 120 - {300 : 226 + 10}
= 120 - \(\dfrac{1280}{113}\)
= \(\dfrac{12280}{113}\)
\(a̸\)
\(\frac{3}{5}-\frac{1}{2}.x=\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{3}{5}-\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{7}{20}\)
\(\Rightarrow\frac{7}{10}\)
\(b̸\)
\(11,3+2\left[x-\frac{1}{3}\right]=\frac{25}{6}\)
\(2\left[x-\frac{1}{3}\right]=\frac{25}{6}-11,3\)
\(2\left[x-\frac{1}{3}\right]=\frac{-107}{15}\)
\(x-\frac{1}{3}=\frac{-107}{15}:2\)
\(x-\frac{1}{3}=\frac{-107}{30}\)
\(x=\frac{-107}{30}+\frac{1}{3}\)
\(x=\frac{-97}{30}\)
\(a)\frac{3}{5}-\frac{1}{2}x=\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{3}{5}-\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{7}{20}\)
\(\implies x=\frac{7}{20}:\frac{1}{2}\)
\(\implies x=\frac{7}{10}\)
Vậy...
\(b) 11,3+2(x-\frac{1}{3})=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-2.\frac{1}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-\frac{2}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x=\frac{25}{6}+\frac{2}{3}\)
\(\implies \frac{113}{10}+2x=\frac{29}{6}\)
\(\implies 2x=\frac{29}{6}-\frac{113}{10}\)
\(\implies 2x=\frac{-97}{15}\)
\(\implies x=\frac{-97}{30}\)
Vậy..
\(c)5x-435+2x+140+3x=565\)
\(\implies (5x+2x+3x)+(-435+140)=565\)
\(\implies 10x+(-295)=565\)
\(\implies 10x=565-(-295)\)
\(\implies 10x=860\)
\(\implies x=86\)
Vậy...
~ hok tốt a~
\(5x^2-15x-140=0\)
Ta có \(\Delta=15^2+4.5.140=3025,\sqrt{\Delta}=55\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{15+55}{10}=7\\x=\frac{15-55}{10}=-4\end{cases}}\)
Bài làm
5x² - 15x - 140 = 0
<=> 5x² + 35x - 20x - 140 = 0
<=> 5x( x + 7 ) - 20( x - 7 ) = 0
<=> ( x - 7 )( 5x - 20 ) = 0
<=> x - 7 = 0 hoặc 5x - 20 = 0
<=> x = 7 hoặc x = 4
Vậy S = { 7;4}
\(\left(\frac{2}{5}x:10\right)\cdot\left(\frac{3}{5}x:15\right)=4\)
\(\Leftrightarrow\left(\frac{2}{5}x\cdot\frac{1}{10}\right)\cdot\left(\frac{3}{5}x\cdot\frac{1}{15}\right)=4\)
\(\Leftrightarrow\frac{1}{25}x\cdot\frac{1}{25}x=4\)
\(\Leftrightarrow x\left(\frac{1}{25}+\frac{1}{25}\right)=4\)
\(\Leftrightarrow x\cdot\frac{2}{25}=4\)
\(\Leftrightarrow x=50\)
(2/5x : 10) x ( 3/5x : 15) = 4
(2/5x x 1/10) x ( 1/5x x 1/15) = 4
1/ 25x x 1/25x = 4
1/625x2 = 4
x2 = 4 : 1/625
x2 = 2500
x = 50
140:(5.X-15)=4
5.X-15=140:4
5.X-15=35
5.X=35+15
5.X=50
X=50:5
=>X=10