a, (2x-3)2=16
b, (3x-2)5=-243
c, (x-7)x+1=(x-7)x+11
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LL
27 tháng 6 2019
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
27 tháng 6 2019
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
12 tháng 6 2019
a) \(2x\left(x-3\right)-x\left(2x+1\right)-3\left(x+5\right)=11\)
\(\Rightarrow2x^2-6x-2x^2-x-3x-15=11\)
\(\Rightarrow-10x=26\Rightarrow x=-2,6\)
Vậy ...........
b) \(x\left(x-1\right)-\left(x^2+3x-5\right)-2\left(x+3\right)=17\)
\(\Rightarrow x^2-x-x^2-3x+5-2x-6=17\)
\(\Rightarrow-6x=18\Rightarrow x=-3\)
c) \(5x\left(x-7\right)-\left(5x+1\right)x-\left(x+3\right)2=13\)
\(\Rightarrow5x^2-35x-5x^2-x-2x-6=13\)
\(\Rightarrow-38x=19\Rightarrow x=-\frac{1}{2}\)
d) \(\left(2x^2-3x+5\right)-2x\left(x-3\right)+\left(x-1\right)\left(-2\right)=10\)
\(\Rightarrow2x^2-3x+5-2x^2+6x-2x+2=10\)
\(\Rightarrow x=3\)
24 tháng 9 2021
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
17 tháng 5 2022
a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
19 tháng 6 2022
a: \(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)
=>13/12x=13/12
hay x=1
b: \(\Leftrightarrow\dfrac{3x-11}{11}-\dfrac{x}{3}=\dfrac{3x-5}{7}-\dfrac{5x-3}{9}\)
\(\Leftrightarrow\dfrac{3}{11}x-1-\dfrac{1}{3}x=\dfrac{3}{7}x-\dfrac{5}{7}-\dfrac{5}{9}x+\dfrac{1}{3}\)
\(\Leftrightarrow x\cdot\dfrac{46}{693}=\dfrac{13}{21}\)
hay x=429/46
9 tháng 6 2016
a)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x-3x^2+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x-1=\frac{5}{2}\Leftrightarrow2x=1+\frac{5}{2}=\frac{7}{2}\Leftrightarrow x=\frac{7}{4}\)
b)
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow8x+4-4x+1+8=11\Leftrightarrow4x+13=11\Leftrightarrow4x=-2\Leftrightarrow x=-\frac{1}{2}\)
c)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow-4x+1+6x+9+245=0\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{255}{2}\).
9 tháng 6 2016
a ) ( 3x2 - x + 1 ) ( x + 1 ) + x2 ( 4 - 3x ) = 5/2
=> 3x3 + 3x2 - x2 - x + x + 1 + 4x2 - 3x3 = 5/2
=> 6x2 + 1 = 5/2
=> 6x2 = 1,5
=> x2 = 0,25
=> x = 0,5
24 tháng 9 2017
Bài4:
\(a,A=2x\left(x-1\right)-x\left(2x+1\right)-\left(3-3x\right)\\ =2x^2-2x-2x^2-x-3+3x\\ =-3\)
Vậy...(đpcm)
\(b,B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\\ =2x^2-6x-2x^2+6x-4\\ =-4\)
Vậy...(đpcm)
\(c,C=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\\ =6x^2+23x-55-6x^2-23x-21\\ =-21\)
Vậy...(đpcm)
d,Câu d là câu c
24 tháng 9 2017
\(A=2x\left(x-1\right)-x\left(2x+1\right)-\left(3-3x\right)\)
\(A=2x^2-2x-2x^2-x-3+3x\)
\(A=-3\)
\(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)
\(B=2x^2-6x-2x^2+4x-2x+4x-4\)
\(B=-4\)
22 tháng 5 2022
a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)
Do đó: A=C+D
\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-4x^2-12x-9-5+20x\)
\(=-30\)
\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)
\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
=-175
A=C+D=-30-175=-205
b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)
\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
Do đó: B=E+F
\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)
\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)
\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)
\(=-18x^3+8x^2+7x-4\)
\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)
\(=-95\)
\(B=-18x^3+8x^2+7x-99\)
a) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(2x-3=4\)
\(2x=7\)
\(x=\dfrac{7}{2}=3,5\)
b) \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=-3^5\)
\(3x-2=-3\)
\(3x=-1\)
\(3x=-\dfrac{1}{3}\)
c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)
\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)
\(x-7=0;\left(x-7\right)^{10}=1\)
\(x=7;\left(x-7=1;x-7=-1\right)\)
\(x=7;x=8;x=6\)
a, (2\(x\) - 3)2 = 16
\(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}
b, (3\(x\) - 2)5 = -243
( 3\(x\) - 2)5 = (-3)5
3\(x\) - 2 = -3
3 \(x\) = -1
\(x\) = - \(\dfrac{1}{3}\)
Vậy \(x\) = -\(\dfrac{1}{3}\)
c, \(\left(x-7\right)\)\(x+1\) = (\(x-7\))\(x+11\)
(\(x-7\))\(^{x+1}\).( \(\left(x-7\right)^{10}\) - 1 ) = 0
\(\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)
Vậy \(x\in\){ 6; 7; 8}