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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
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a.Ta có: 2|x| + 3|1-x| - |x-3| =0
2x + 3 - 3x - x +3 =0
-2x + 6 =0
-2x =-6
x =3
b.Ta có:4|3x-1| + |x| - 2|x-5| +7|x-3| =12
12x-4 + x -2(x-5) + 7x - 3 =12
12x-4+ x - 2x +10 + 7x - 3=12
18x + 3 =12
18x =12-3
18x =9
x =1/2
a. 2 |x |+ 3 | 1- x| - 5 | x - 3= 0
2 x + 3 - 3 x - x + 3 = 0
-2x + 6 = 0
- 2 x = - 6
x = 3
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Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
TH1:x < 0 . PT có dạng
\(-4\left(3x-1\right)-x+2\left(x-5\right)-7\left(x-3\right)=12\)
\(4-12x-x+2x-10-7x+21=12\)
\(15-18x=12\)
\(x=\frac{1}{6}\left(koTM\right)\)
TH2:\(0\le x\le\frac{1}{3}\) PT có dạng:
\(x-4\left(3x-1\right)-7\left(x-3\right)+2\left(x-5\right)=12\)
\(x=\frac{3}{16}\left(TM\right)\)
TH3:\(\frac{1}{3}\le x< 3\) PT có dạng:
\(x+4\left(3x-1\right)-7\left(x-3\right)+2\left(x-5\right)=12\)
\(x=\frac{5}{8}\left(TM\right)\)
TH4:\(3\le x< 5\) PT có dạng:
\(4\left(3x-1\right)+x+2\left(x-5\right)+7\left(x-3\right)=12\)
\(x=\frac{47}{22}\left(koTM\right)\)
\(TH5:x\ge5\)PT có dạng:
\(4\left(3x-1\right)+x-2\left(x-5\right)+7\left(x-3\right)=12\)
\(x=1,5\left(koTM\right)\)
Vậy nghiệm PT là \(\frac{3}{16};\frac{5}{8}\)