Tìm x: 35 - [ ( 2x - 3)2 : 7 ] = 28
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( x - 1 ) + 5 = 17
x - 1 = 17 - 5
x - 1 = 12
x = 12 + 1
x = 13
\(-(35-x)-(37-x)=33\\\Rightarrow-35+x-37+x=33\\\Rightarrow(-35-37)+(x+x)=33\\\Rightarrow-72+2x=33\\\Rightarrow2x=33+72\\\Rightarrow2x=105\\\Rightarrow x=105:2\\\Rightarrow x=52,5\\---\)
\((-14)+x-7=(-10)\\\Rightarrow x-7=(-10)+14\\\Rightarrow x-7=4\\\Rightarrow x=4+7\\\Rightarrow x=11\)
-(35 - \(x\)) - (37 - \(x\)) = 33
- 35 + \(x\) - 37 + \(x\) = 33
(\(x\) + \(x\)) - (35 + 37) = 33
2\(x\) - 72 = 33
2\(x\) = 33 + 72
2\(x\) = 105
\(x\) = \(\dfrac{105}{2}\)
(-14) + \(x\) - 7 = (-10)
-14 + \(x\) - 7 = -10
\(x\) - (14 + 7) = -10
\(x\) - 21 = -10
\(x\) = -10 + 21
\(x\) = 11
Ta có:\(\left(4x+28\right)⋮\left(2x+1\right)\Rightarrow\left[\left(4x+2\right)+26\right]⋮\left[2\left(2x+1\right)\right]\)
\(\Rightarrow\left[\left(4x+2\right)+26\right]⋮\left(4x+2\right)\)
\(\Rightarrow26⋮\left(4x+2\right)\)(t/c chia hết của 1 tổng)
Vì \(x\in N\Rightarrow4x+2\in N\)(1)
\(\Rightarrow\left(4x+2\right)\in\left\{2;13;26\right\}\)
\(\Rightarrow x\in\left\{0;\frac{11}{4};6\right\}\)
Từ (1)=>\(x\in\left\{0;6\right\}\)
Có gì ko hiểu thì kbạn với mình nha
\(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{21}{35}+\dfrac{20}{35}\\ \Rightarrow x=\dfrac{41}{35}\)
Vậy `x=41/35`
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\(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{3}\right)x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{9}{21}-\dfrac{14}{21}\right)x=\dfrac{10}{21}\\ \Rightarrow\dfrac{-5}{21}x=\dfrac{10}{21}\\ \Rightarrow x=\dfrac{10}{21}:\left(-\dfrac{5}{21}\right)\\ \Rightarrow x=-2\)
Vậy `x=-2`
a)
-4/7 - x = 3/5 - 2x
2x - x = 3/5 + 4/7
x = 41/35
Vậy x = 41/35
b)
3/7.x - 2/3.x = 10/21
x(3/7 - 2/3) = 10/21
x.(-5/21) = 10/21
x = 10/21 : (-5/21) = -2
Vậy x = -2
\(\frac{x}{7}=\frac{x+36}{35}\)
\(\Rightarrow x.35=7.\left(x+36\right)\)
\(\Rightarrow35x=7x+252\)
\(\Rightarrow35x-7x=252\)
\(\Rightarrow28x=252\)
\(\Rightarrow x=9\)
a) \(\frac{x}{y}=\frac{5}{7}\)=>\(\frac{x}{5}=\frac{y}{7}=>\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\frac{xy}{5.7}\)
=>\(\frac{x^2}{25}=\frac{y^2}{49}=\frac{35}{35}=1\)
=> \(x^2=25;y^2=49\)
=>\(x=\pm5;y=\pm7\)
\(35-\left[\left(2x-3\right)^2:7\right]=28\)
\(\Rightarrow\left[\left(2x-3\right)^2:7\right]=35-28\)
\(\Rightarrow\left(2x-3\right)^2:7=7\)
\(\Rightarrow\left(2x-3\right)^2=1\)
\(\Rightarrow2x-3=\pm1\)
\(\Rightarrow x=2\) hay \(x=1\)
35 - [(2\(x\) - 3)2:7 ] = 28
(2\(x-3\))2 : 7 = 35 - 28
(2\(x\) - 3)2 : 7 = 7
(2\(x\) - 3)2 = 7 \(\times\) 7
(2\(x-3\))2 = 72
\(\left[{}\begin{matrix}2x-3=-7\\2x-3=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-7+3\\2x=7+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy \(x\in\) {-2; 5}