(x-140);7=33-23*3
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140-x/x-20=140/20
140-x/x-20=7
140-x/x=7+20
140-x/x=27
x/x=140-27
x/x=113/1
\(\dfrac{140}{x}+5=\dfrac{\left(140+10\right)}{x-1}\left(x\ne0,x\ne1\right)\)
\(\Leftrightarrow\dfrac{140+5x}{x}=\dfrac{150}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(140+5x\right)=150x\)
\(\Leftrightarrow140x+5x^2-140-5x-150x=0\)
\(\Leftrightarrow5x^2-15x-140=0\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=-4\left(N\right)\end{matrix}\right.\)
\(S=\left\{7,-4\right\}\)
ĐK: `x \ne 0 ; x \ne -1`
`140/x+5=150/(x-1)`
`<=>(140+5x)/x=150/(x-1)`
`<=>(140x+5x)(x-1)=150x`
`<=>5x^2+135x-140=150x`
`<=>5x^2-15x-140=0`
`<=>` \(\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy...
\(\dfrac{140}{x}-\dfrac{140}{x-6}=\dfrac{-1}{6}\)
\(\Leftrightarrow\dfrac{140\left(x-6\right)-140x}{x\left(x-6\right)}=-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{-840}{x\left(x-6\right)}=-\dfrac{1}{6}\)
\(\Leftrightarrow x^2-6x-5040=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=74.05\\x=-68.05\end{matrix}\right.\)
Lời giải:
\(\frac{140}{x}-\frac{140}{x-6}=\frac{-1}{6}\Leftrightarrow \frac{140(x-6-x)}{x(x-6)}=\frac{-1}{6}\)
\(\Leftrightarrow x(x-6)=5040\Leftrightarrow x^2-6x-5040=0\)
\(\Leftrightarrow (x-3)^2-5049=0\)
\(\Leftrightarrow (x-3-\sqrt{5049})(x-3+\sqrt{5049})=0\Rightarrow x=3\pm \sqrt{5049}\)
a, x - 140 : 35= 270
x - 4 = 270
x = 274
b, (x - 140) : 35= 270
x - 140 = 9450
x = 9590
137+4 (2x+3)=200(hình như đề bài sai)
200-6(x-5)=140
6(x-5)=200-140
6(x-5)=60
x-5=60:6
x-5=10
x=10+5
x=15
140:(x-8)=7
x-8=140:7
x-8=20
x=20+8
x=28
137+4(2x+3)=200
4(2x+3)=200-137
4(2x+3)=63
2x+3=63:4
2x+3=\(\frac{63}{4}\)
2x=\(\frac{63}{4}\)-3
2x=\(\frac{51}{4}\)
x=\(\frac{51}{4}\):2
x=\(\frac{51}{8}\)
Vậy ...
200-6(x-5)=140
6(x-5)=200-140
6(x-5)=60
x-5=60:6
x-5=10
x=15
Vậy...
140:(x-8)=7
x-8=140:7
x-8=20
x=20+8
x=28
Vậy...
k mik nhé
\(140\%\times x+0,5\times x=0,75\\ \Rightarrow1,4\times x+0,5\times x=0,75\\ \Rightarrow\left(1,4+0,5\right)\times x=0,75\\ \Rightarrow1,9\times x=0,75\\ \Rightarrow x=0,75:1,9\\ \Rightarrow x=\dfrac{15}{38}\)
\(\dfrac{140}{100}\times x+\dfrac{1}{2}\times x=0,75\\ x\times\left(\dfrac{140}{100}+\dfrac{1}{2}\right)=0,75\\ x\times\dfrac{19}{10}=0,75\\ x=\dfrac{15}{38}\)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+140\right)=5041\)
\(x+x+...+x+2+4+...+140=5041\)
Có tất cả số hạng là:
\(\dfrac{\left(140-2\right)}{2}+1=70\left(số\right)\)
=> \(71x+\dfrac{\left(140+2\right).70}{2}=5041\)
=> \(71x=71\)
=> \(x=1\)
x + (x + 2) + (x + 4) + ... + (x + 140) = 5041
x + 70x + (140 + 2) . 70 : 2 = 5041
71x + 4970 = 5041
71x = 5041 - 4970
71x = 71
x = 71 : 71
x = 1
\(\left(x+140\right):x+260\text{=}331\)
\(\left(x+140\right):x\text{=}331-260\)
\(x:x+140:x\text{=}71\)
\(1+140:x\text{=}71\)
\(140:x\text{=}71-1\)
\(140:x\text{=}70\)
\(x\text{=}140:70\)
\(x\text{=}2\)
Trước tiên, bạn cần biết về tình chất phân số nhé:
\(\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\)
Áp dụng vào bài, ta có:
\(\dfrac{\left(x+140\right)}{x}+260=331\)
\(\Rightarrow\dfrac{x}{x}+\dfrac{40}{x}+260=331\)
\(\Rightarrow1+\dfrac{40}{x}+260=331\)
\(\Rightarrow40:x=331-1-260=70\)
\(\Rightarrow x=\dfrac{40}{70}=\dfrac{4}{7}\)