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Tim x, bt:
a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)
b) 5.(3x+5) - 4.(2x-3) =5x + 3x(2x-12) +1
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a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
-20x - 36x - 30x + 6x = -240 - 84 - 72 -84
-80x = -480
x= 6
4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6.(x+14)
4.18 - 4.5x - 12.3x + 12.7 = 15.2x - 15.16 - 6x - 6.14
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
72 + 84 - 20x - 36x = 30x - 6x - 240 - 84
156 - 56x = 24x - 324
156 = 24x - 324 + 56x
156 = 80x - 324
80x - 324 = 156
80x = 156 + 324
80x = 480
x = 480:80
x = 6
câu b giải tương tự
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Ta có A = |x - 2015| + |x - 2016|
= |x - 2015| + |2016 - x|
\(\ge\)|x - 2015 + 2016 - x| = 1
Dấu "=" xảy ra <=> \(\left(x-2015\right)\left(2016-x\right)\ge0\)
TH1 : \(\hept{\begin{cases}x-2015\ge0\\2016-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2015\\x\le2016\end{cases}}\Rightarrow2015\le x\le2016\)
TH2 : \(\hept{\begin{cases}x-2015\le0\\2016-x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2015\\x\ge2016\end{cases}}\left(\text{loại}\right)\)
Vậy Min A = 1 <=> \(2015\le x\le2016\)
b) Ta có B = |x - 5| + |x - 7|+ |2x - 18|
= |x - 5| + |x - 7|+ |18 - 2x|
\(\ge\)|x - 5 + x - 7| + |18 - 2x|
= |2x - 12| + |18 - 2x|
\(\ge\)|2x - 12 + 18 - 2x| = 6
Dấu "=" xảy ra <=> \(\left(2x-12\right)\left(18-2x\right)\ge0\)
TH1 : \(\hept{\begin{cases}2x-12\ge0\\18-2x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge6\\x\le9\end{cases}}\Rightarrow6\le x\le9\)
TH2 : \(\hept{\begin{cases}2x-12\le0\\18-2x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le6\\x\ge9\end{cases}}\)(loại)
Vậy Min B = 6 <=> \(6\le x\le9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{18-x}{5}+\frac{17-x}{6}=\frac{16-x}{7}+\frac{15-x}{8}\)
\(\Leftrightarrow\left(\frac{18-x}{5}+1\right)+\left(\frac{17-x}{6}+1\right)=\left(\frac{16-x}{7}+1\right)+\left(\frac{15-x}{8}+1\right)\)
\(\Leftrightarrow\left(\frac{18-x+5}{5}\right)+\left(\frac{17-x+6}{6}\right)=\left(\frac{16-x+7}{7}\right)+\left(\frac{15-x+8}{8}\right)\)
\(\Leftrightarrow\frac{23-x}{5}+\frac{23-x}{6}=\frac{23-x}{7}+\frac{23-x}{8}\)
\(\Leftrightarrow\frac{23-x}{5}+\frac{23-x}{6}-\frac{23-x}{7}-\frac{23-x}{8}=0\)
\(\Leftrightarrow\left(23-x\right).\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\)
Vì \(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\ne0.\)
\(\Leftrightarrow23-x=0\)
\(\Leftrightarrow x=23-0\)
\(\Leftrightarrow x=23.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{23\right\}.\)
Chúc bạn học tốt!
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a =>5x(x2-6x+9)-5(x3-3x2+3x-1)+15(x2-4)=5
=>5x3-30x2+45x-5x3+15x2+15x+5+152-50=5
=>60x-55=5
=>x=1
c) x2 ( x2 +1 ) - x2 -1 =0
x2 (x2 +1) -(x2 +1) =0
(x2 +1)(x2 -1) =0
*) x2 = -1 --> x không có giá trị thỏa mãn
*) x2 = 1 --> x = 1
Vậy x= 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : 6x(3x + 5) - 2x(9x - 2) + (17 - x)(x - 1) + x(x - 18) = 0
<=> 18x2 + 30x - 18x2 + 4x + 17x - 17 - x2 + x + x2 - 18x = 0
<=> 34x - 17 = 0
<=> 34x = 17
=> x = 2
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Câu B đây;vừa bị lag
B, \(\frac{x+1}{35}\)+\(\frac{x+3}{33}\)=\(\frac{x+5}{31}\)+\(\frac{x+7}{29}\)
⇔ \(\frac{x+1}{35}\)+1+\(\frac{x+3}{33}\)+1=\(\frac{x+5}{31}\)+1+\(\frac{x+7}{29}\)+1
⇔ \(\frac{x+36}{35}\)+\(\frac{x+36}{33}\)-\(\frac{x+36}{31}\)-\(\frac{x+36}{29}\)=0
⇔ (x+36)(\(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\))=0
Mà \(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\)<0
⇔ x+36=0
⇔ x=-36
Vậy tập nghiệm của phương trình đã cho là:S={-36}
câu C tương tự nhé
0