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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
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
$(1-2x)\cdot3-4(6x-1)=7x-5$
$3-6x-24x+4=7x-5$
$7-30x=7x-5$
$12=37x$
$x=\frac{12}{37}$
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 \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
c) \(5x-7=3x+9\)
d) \(5x-\left|9-7x\right|=3\)
e) \(-5+\left|3x-1\right|+6=\left|-4\right|\)
h) \(5^{-1}.25^x=125\)
\(\Rightarrow\frac{1}{5}.25^x=125\)
\(\Rightarrow25^x=125:\frac{1}{5}\)
\(\Rightarrow25^x=625\)
\(\Rightarrow25^x=25^2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
g) \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.\left[1-\left(x-1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=1+1\\x=\left(-1\right)+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{1;2;0\right\}.\)
i) \(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\\left|x+3\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\) \(\forall x.\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0.\)
Lúc này ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow x+1+x+2+x+3=4x\)
\(\Rightarrow\left(x+x+x\right)+\left(1+2+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow6=4x-3x\)
\(\Rightarrow6=1x\)
\(\Rightarrow x=6\left(TM\right).\)
Vậy \(x=6.\)
Chúc bạn học tốt!
c) \(\left(3x-1\right).\left(2x+7\right)-\left(x+1\right).\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=x+2-x+5\)
\(\Leftrightarrow18x-2-7=0\)
\(\Rightarrow x=\dfrac{9}{18}=\dfrac{1}{2}\)
b) \(2.\left(3x-1\right).\left(2x+5\right)-6.\left(2x-1\right).\left(x+2\right)=1\)
\(\Leftrightarrow\left(6x-2\right).\left(2x+5\right)-\left(12x-6\right).\left(x+2\right)=1\)
\(\Leftrightarrow12x^2+30x-4x-10-\left(12x^2+24x-6x-12\right)=1\)
\(\Leftrightarrow12x^2+26x-10-12x^2-18x +12=1\)
\(\Leftrightarrow8x+2=1\)
\(\Rightarrow x=\dfrac{-1}{8}\)
Bài 1:
a) -6x + 3(7 + 2x)
= -6x + 21 + 6x
= (-6x + 6x) + 21
= 21
b) 15y - 5(6x + 3y)
= 15y - 30 - 15y
= (15y - 15y) - 30
= -30
c) x(2x + 1) - x2(x + 2) + (x3 - x + 3)
= 2x2 + x - x3 - 2x2 + x3 - x + 3
= (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
= 3
d) x(5x - 4)3x2(x - 1) ??? :V
Bài 2:
a) 3x + 2(5 - x) = 0
<=> 3x + 10 - 2x = 0
<=> x + 10 = 0
<=> x = -10
=> x = -10
b) 3x2 - 3x(-2 + x) = 36
<=> 3x2 + 2x - 3x2 = 36
<=> 6x = 36
<=> x = 6
=> x = 5
c) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
=> x = -2
=>\(\frac{2}{3}x-\frac{6}{7}=\frac{6}{5}x-\frac{1}{6}\)
=>\(\frac{6}{5}x-\frac{2}{3}x=\frac{1}{6}-\frac{6}{7}\)
=>\(\frac{8}{15}x=-\frac{29}{42}\)
=>\(x=\frac{-29}{42}:\frac{8}{15}\)
=>\(x=\frac{-145}{112}\)