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HT
1
NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
HT
2
25 tháng 5 2015
a/3x(12x - 4 ) -9x (4x -3 ) = 30
<=> 36x^2 - 12x - 36x²+27x = 30
<=> 15x = 30
<=> x=2
b/ => 5x -2. x^2 + 2.x^2 -2x = 15
=> 5x -2x = 15
=> 3x = 15 => x= 5
25 tháng 5 2015
a/3x(12x - 4 ) -9x (4x -3 ) = 30
36x^2 - 12x - 36x²+27x = 30
15x = 30
x=2
b/ 5x -2. x^2 + 2.x^2 -2x = 15
5x -2x = 15
3x = 15 => x= 5
C
4
PN
24 tháng 8 2021
a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)
c nhường mấy bn khácccc
24 tháng 8 2021
a) x^2-6x+9=4.
x=1, x=5
b) x^2(x-3)-(4X-12)=0
x=-2, x=2, x=3
c) (2x+3)^2-4(x+2)^2=12
x=-19/4
ES
3
13 tháng 7 2016
\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)
\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)
\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)
\(\Rightarrow-14x-60=0\)
\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)
\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)
\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)
\(\Rightarrow-x^2+12x-32=7x-x^2-10\)
\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)
\(\Rightarrow5x-22=0\)
\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)
PA
13 tháng 7 2016
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
15x + 25 - 8x + 12 = 5x + 16x + 96 + 1
15x - 8x - 5x - 16x = 96 + 1 - 25 - 12
-14x = 60
x = \(\frac{60}{-14}\)
x = \(-\frac{30}{7}\)
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x2 + 2x
(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x2 + 2x
(x - 4)(-x + 8) = 5x - 10 - x2 + 2x
-x2 + 8x + 4x - 32 = 5x - 10 - x2 + 2x
(-x2 + x2) + (8x + 4x - 5x - 2x) = -10 + 32
5x = 22
x = \(\frac{22}{5}\)
16 tháng 7 2020
\(\left(x-3\right)\left(x+x^2\right)+\left(x-5\right)\left(x+1\right)-x^3=12\)
\(\Leftrightarrow x^2+x^3-3x-3x^2+x^2+x-5x-5-x^3=12\)
\(\Leftrightarrow-x^2-7x-5=12\Leftrightarrow-x^2-7x-17=0\)
Ta có : \(\left(-7\right)^2-4\left(-17\right)\left(-1\right)< 0\)Vậy phương trình vô nghiệm
MT
3
19 tháng 9 2016
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\) .Trừ 1 ở mỗi hạng tử của 2 vế ,ta có :
\(\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)\left(x-11\right)=\left(\frac{1}{14}+\frac{1}{15}\right)\left(x-11\right)\)
\(\Rightarrow\left[\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\right]\left(x-11\right)=0\)
\(\frac{1}{12}>\frac{1}{14};\frac{1}{13}>\frac{1}{15}\Rightarrow\frac{1}{12}+\frac{1}{13}>\frac{1}{14}+\frac{1}{15}\Rightarrow\left(\frac{1}{12}+\frac{1}{13}\right)-\left(\frac{1}{14}+\frac{1}{15}\right)\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
MA
19 tháng 9 2016
\(\frac{x+1}{12}+\frac{x+2}{13}=\frac{x+3}{14}+\frac{x+4}{15}\)
\(\Leftrightarrow\frac{x+1}{12}-1+\frac{x+2}{13}-1=\frac{x+3}{14}-1+\frac{x+4}{15}-1\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}=\frac{x-11}{14}+\frac{x-11}{15}\)
\(\Leftrightarrow\frac{x-11}{12}+\frac{x-11}{13}-\frac{x-11}{14}-\frac{x-11}{15}=0\)
\(\Leftrightarrow\left(x-11\right)\left(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)
Mà: \(\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\)
\(\Rightarrow x-11=0\Rightarrow x=11\)
23 tháng 9 2021
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
23 tháng 9 2021
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
TN
10 tháng 7 2018
\(12\left(x-2\right)\left(x+2\right)-3\left(2x+3\right)^2\) \(=52\)
\(12\left(x^2-4\right)-3\left(4x^2+12x+9\right)\) \(=52\)
\(12x^2-48-12x^2-36x-27\) \(=52\)
\(-36x-75=52\)
\(-36x=127\)
\(x=\frac{-127}{36}\)
\(\left(2x+1\right)^2-4\left(x-1\right)\left(x+1\right)\) \(+2x=5\)
\(4x^2+4x+1-4\left(x^2-1\right)\) \(+2x=5\)
\(4x^2+4x-1-4x^2+4+2x=5\)
\(6x+3=5\)
\(6x=2\)
\(x=3\)
\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\) \(+6\left(x-1\right)^2=15\)
\(x^3-6x^2+12x-8-\left(x-3\right)\left(x+3\right)^2\) \(+6\left(x^2-2x+1\right)=15\)
\(x^3-6x^2+12x-8-\left(x^2-9\right)\left(x+3\right)\) \(+6x^2-12x+6=15\)
\(x^3-2\) \(-\left(x^3+3x^2-9x-27\right)\)\(=15\)
\(x^3-2-x^3-3x^2+9x+27=15\)
\(-3x^2+9x+25=15\)
\(-3x^2+9x+10=0\)
\(-3\left(x^2-3x-\frac{10}{3}\right)\) \(=0\)
\(x=\frac{9+\sqrt{201}}{6}\)
các câu còn lại tương tự
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-12=12\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+4=t\)
\(t\left(t+2\right)-24=0\Leftrightarrow t^2+2t-24=0\Leftrightarrow t=4;t=-6\)
Theo cách đặt \(x^2+5x=0\Leftrightarrow x=0;x=-5\)
\(x^2+5x+10=0\Leftrightarrow x=\frac{-5\pm\sqrt{15}i}{2}\)