Tìm x biết
\(|x-5|\)+ 3 * (x-5) = 3x+7
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\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
a) |3x - 5| = 3x - 5 => 3x - 5 > 0 => 3x > 5 => x > 5/3
b) |7 - x| = x - 7 => 7 - x < 0 => - x < - 7 => x > 7
c) |2x - 3| = 3 - 2x => 2x - 3 < 0 => 2x < 3 => x < 3/2
a) ( 2x - 3 ) - ( x - 5 ) = ( x + 7 ) - ( x + 2 )
<=> 2x - 3 - x + 5 = x + 7 - x - 2
<=> x = 3
b)(7x-5)-(6x+4)=(2x+3)-(2x+1)
<=> 7x - 5 - 6x - 4 = 2x + 3 - 2x - 1
<=> x = 11
c)(9x-3)-(8x+5)=(3x+2)
<=> 9x - 3 - 8x - 5 = 3x + 2
<=> -2x = 10
<=> x = -5
d)(x+7)-(2x+3)=(3x+5)-(2x+4)
<=> x + 7 - 2x - 3 = 3x + 5 - 2x - 4
<=> -2x = -3
<=> x = 3/2
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Tìm x biết:
5. ( x-1 ) - 7.( x-2 ) = 2x -39
Tìm x thuộc Z biết:
x - 3 - 14.( x-2 )= -3x -3
\(3x+7⋮x-2\)
5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39
<=> 5x - 5 - 7x + 14 = 2x - 39
<=> 5x - 7x - 2x = -39 + 5 - 14
<=> -4x = -48
<=> x = 12
x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)
\(\Rightarrow\chi-3-28+3=-3\chi-3\)
\(\Rightarrow\chi-28=11\chi\)
\(\Rightarrow\chi-11\chi=28\)
\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\)
a)4×(x-5)-(x-1)×(4x-3)=5
=>4x-20-4x2+7x-3-5=0
=>-4x2+11x-28=0
=>-4(x2-\(\frac{11x}{4}\)+7)=0
=>\(-4\left(x-\frac{11}{8}\right)^2-\frac{327}{16}< 0\)
=>vô nghiệm
b) (3x-4)(x-2)=3x(x-9)-3
=>3x2-10x+8=3x2-27x-3
=>17x=-11
=>x=-11/17
c)(x-5)×(x-4)-(x+1)×(x-2)=7
=>x2-9x+20-x2+x+2=7
=>22-8x=7
=>-8x=-15
=>x=8/15
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)