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Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a, x=-505
b, x=35/8 hoac -37/8
nhung cau con lai thi tong tu
|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
Ai chẳng biết chuyển vế đổi dấu :v
a) \(x-7=4x+10\)
\(x-4x=10+7\)
\(-3x=17\)
\(x=\dfrac{17}{-3}\)
Vậy \(x=\dfrac{17}{-3}\)
b) \(2x+5=-3x+7\)
\(2x+3x=7-5\)
\(5x=2\)
\(x=\dfrac{2}{5}\)
Vậy \(x=\dfrac{2}{5}\)
c) \(x-\left(3x+7\right)=6x-1\)
\(x-3x-7=6x-1\)
\(-2x-7=6x+1\)
\(-7-1=6x+2x\)
\(-8=8x\)
\(x=\dfrac{-8}{8}=-1\)
Vậy \(x=-1\)
d) \(x+\left(5x-1\right)=15\)
\(x+5x-1=15\)
\(6x=15+1\)
\(6x=16\)
\(x=\dfrac{16}{6}=\dfrac{8}{3}\)
Vậy \(x=\dfrac{8}{3}\)
1 , x - 7 = 4x + 10
x - 4x = 10 + 7
- 3x = 17
x = 17 : ( - 3 )
x = \(\dfrac{-17}{3}\)
2 , 2x + 5 = -3x + 7
2x + 3x = 7 -5
5x = 2
x = 2 : 5
x =\(\dfrac{2}{5}\)
3 , x - ( 3x + 7 ) = 6x - 1
x - 3x - 7 = 6x - 1
x - 3x -6x = -1 +7
-8x = 6
x = 6 : ( -8 )
x = \(\dfrac{-3}{4}\)
4 , x + ( 5x -1 ) = 15
x + 5x - 1 = 15
x + 5x = 15 + 1
6x = 16
x = 16 : 6
x = \(\dfrac{8}{3}\)
5 , / x + 1 / = / 2x - 5 /
TH 1 : x + 1 = 2x - 5
x - 2x = -5 -1
- x = -4
= > x = 4
TH 2 : -x -1 = -2x + 5
-x + 2x = 5 + 1
x = 6
6 , / 3x + 8 / - / x -10 / = 0
3x + 8 - x + 10 = 0
3x - x = 0 - 10 - 8
2 x = -18
x = -18 : 2
x = - 9
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
a, \(\dfrac{x}{2}+\dfrac{3x}{5}=-\dfrac{3}{2}\Rightarrow5x+6x=-15\Leftrightarrow x=-\dfrac{15}{11}\)
b, TH1 : \(\dfrac{2}{3}x-\dfrac{4}{7}=0\Leftrightarrow x=\dfrac{6}{7}\);TH2 : \(\dfrac{1}{2}-\dfrac{3}{7x}=0\Rightarrow7x-6=0\Leftrightarrow x=\dfrac{6}{7}\)
c, TH1 : \(\dfrac{4}{5}-2x=0\Leftrightarrow x=\dfrac{4}{5}:2=\dfrac{2}{5}\)
TH2 : \(\dfrac{1}{3}+\dfrac{3}{5x}=0\Rightarrow5x+9=0\Leftrightarrow x=-\dfrac{9}{5}\)