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\(3x\left(x-2\right)-x+2=0\)
\(\Leftrightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
\(B1:\)
\(3x\left(x-2\right)-\left(x-2\right)=0\)
\(\left(3x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(3x - 1)2 - 16 = 0
<=> (3x - 1)2 - 42 = 0
<=> (3x - 1 - 4)(3x - 1 + 4) = 0
<=> (3x - 5)(3x + 3) = 0
<=> 3x - 5 = 0 hoặc 3x + 3 = 0
<=> x = 5/3 hoặc x = - 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^3+2x^2+2x+3=0\)
\(\Leftrightarrow3\left(x^3+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow3\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2-x+3\right)=0\)
Mà \(3x^2-x+3=3\left[\left(x-\frac{1}{6}\right)^2+\frac{35}{36}\right]>0\forall x\)
Do đó: \(x+1=0\Leftrightarrow x=-1\)
Tập nghiệm: \(S=\left\{-1\right\}\)
\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
\(\Leftrightarrow\left[\left(x-1\right)+\left(2x+3\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=27x^3+8\)
\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9\right)=27x^3+8\)
\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(6x^2-15x-9\right)=0\)(Chuyển vế)
\(\Leftrightarrow3\left(3x+2\right)\left(2x^2-5x-3\right)=0\)
\(\Leftrightarrow3\left(3x+2\right)\left(x-3\right)\left(2x+1\right)=0\)
Tập nghiệm: \(S=\left\{-\frac{2}{3};3;-\frac{1}{2}\right\}\)
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a, 3x^2-27x=0
3x(x-9)=0
3x=0=>x=0
x-9=0=>x=9
b,2/3x(x^2-4)=0
2/3x=0=>x=0
x^2-4=0=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2-27x=0\)
\(3x\left(x-9\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-9=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)
\(\frac{2}{3}x\left(x^2-4\right)=0\)
\(\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+2=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)
a)\(3x^2-27x=0\)
\(3x\left(x-9\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-9=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)
b) \(\frac{2}{3}x\left(x^2-4\right)=0\)
\(\frac{2}{3}x\left(x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\frac{2}{3}x=0\\x+2=0\\x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=2\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x^2-27x-64=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{27}{2}+\frac{729}{4}-\frac{985}{4}=0\)
\(\Leftrightarrow\left(x-\frac{27}{2}\right)^2=\frac{985}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{27}{2}=\sqrt{\frac{985}{4}}\\x-\frac{27}{2}=-\sqrt{\frac{985}{4}}\end{matrix}\right.\)\(\left[{}\begin{matrix}x=\sqrt{\frac{985}{4}}+\frac{27}{2}=\frac{27+\sqrt{985}}{2}\\x=-\sqrt{\frac{985}{4}}+\frac{27}{2}=\frac{27-\sqrt{985}}{2}\end{matrix}\right.\)
Vậy: \(x=\frac{27\pm\sqrt{985}}{2}\)
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1. a) 1012 - 992 = (101 + 99)(101 - 99) = 200 . 2 = 400
b) 98.102 = (100 - 2)(100 + 2) = 1002 - 4 = 10000 - 4 = 9996
c) 772 + 232 + 77.46 = 772 + 232 + 77.23.2 = (23 + 77)2 = 1002 = 10000
d) M = x3 + 9x2 + 27x + 27 = (x + 3)3 = (7 + 3)3 = 103 = 1000
2. a) 2x2 + 3x - 5 = 0
=> 2x2 + 5x - 2x - 5 = 0
=> x(2x + 5) - (2x + 5) = 0
=> (x - 1)(2x + 5) = 0
=> \(\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
b) 2x2 - 11x - 51 = 0
=> 2x2 - 17x + 6x - 51 = 0
=> x(2x - 17) + 3(2x - 17) = 0
=> (x + 3)(2x - 17) = 0
=> \(\orbr{\begin{cases}x+3=0\\2x-17=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3\\x=\frac{17}{2}\end{cases}}\)
a) 1012 - 992 = (101-99)(101+99)= 2,200 = 4002
b)98.102 = (100-2)(100+2) = 1002 - 22 =10000 - 4 = 9996
c) 772 + 232 +77.46 = 772 + 232 +2.77.23 = ( 77+23)2 = 1002 =1000
d) Với x=7 => M = 73+ 9.73 + 27.7 + 27 = 10.73 +27.8 = 10.343 + 216 = 3430+216 = 3646
2. a) 2x2 + 3x -5 =0
=> 2(x2 +3/2 x +9/16) -49/8 = 0
=> 2 (x+3/4)2 =49/8
=> (x+3/4)2 =49/16 = (7/4)2 = (-7/4)2
=> x+3/4 = 7/4 hoặc x+3/4 = -7/4
=> x= 1 hoặc x=-5/2
b) 2x2 -11x - 51 =0
=> 2(x2 -11/2x + 121/16) -529/8 = 0
=> (x -11/4)2 = 529/16 = (23/4)2 =(-23/4)2
=> x-11/4=23/4 hoặc x-11/4 = -23/4
=> x=17/2 hoặc x=-3
3x2-27x=0
<=> 3x(x-9)=0
\(\Rightarrow\orbr{\begin{cases}3x=0\\x-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=9\end{cases}}}\)
Vậy x=0 hoặc x=9
Theo đầu bài ta thấy :
\(3x^2=27x\)( vì 2 số giống nhau trừ đi nhau bằng 0 )
\(x^2:x=27:3\)
\(x=9\)
Vậy x = 9