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a) \(\left(x+2\right)^2-9=0\)
\(=>\left(x+2\right)^2-3^2=0\\ =>\left(x+2-3\right).\left(x+2+3\right)=0\)
\(=>\left(x-1\right).\left(x+5\right)=0\)
\(=>\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}=>\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)
Vậy x= 1 hoặc x= -5
b) \(x^2-2x+1=25\)
\(=>x^2-2.x.x+1^2=25\)
\(=>\left(x-1\right)^2-25=0\\ =>\left(x-1\right)^2-5^2=0\)
\(=>\left(x-1-5\right).\left(x-1+5\right)=0\)
\(=>\left(x-6\right).\left(x+4\right)=0=>\orbr{\begin{cases}x-6=0\\x+4=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Vậy x= 6 hoặc x= -4
c) \(4x\left(x-1\right)-\left(2x+5\right)\left(2x-5\right)=1\)
\(=>4x\left(x-1\right)-\left[\left(2x\right)^2-5^2\right]=1\)
\(=>4x\left(x-1\right)-4x^2+25-1=0\)
\(=>4x\left(x-1\right)-4x^2+24=0\)
\(=>4x\left(x-1\right)-\left(4x^2-24\right)=0\\ =>4x\left(x-1\right)-4\left(x^2-6\right)=0\)
..................... tắc ròi -.-"
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+3\right)=15\)
\(=>x^3+27-x^3-3x=15\)
\(=>27-3x-15=0=>12-3x=0=>3\left(4-x\right)=0\)
Vì \(3>0=>4-x=0=>x=4\)
Vậy x= 4
e) \(3\left(x+2\right)^2+\left(2x+1\right)^2-7\left(x+3\right)\left(x-3\right)=28\)
\(=>3\left(x^2+2.x.2+2^2\right)+4x^2+4x+1-7\left(x^2-9\right)=28\)
\(=>3\left(x^2+4x+4\right)+4x^2+4x+1-7x^2+63=28\)
\(=>3x^2+12x+12+4x^2+4x+1-7x^2+63=28\)
\(=>16x+75=28=>16x=-47=>x=\frac{-47}{16}\)
Cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>'-'
Cho phép mình làm câu b trước :v
b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Rightarrow x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)
\(\Leftrightarrow-3x^2\cdot\left(x^2-1\right)=0\)
\(\Leftrightarrow x^2\cdot\left(x^2-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2=0\\x^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1;x=-1\end{matrix}\right.\)
Vậy \(x_1=-1;x_2=0;x=1\)
Xử câu a nào :D
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Rightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow0+3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x+26=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\3x=-26\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
Vậy \(x_1=0;x_2=-\dfrac{26}{3}\)
Ok con tê tê 
b)(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49
(=) x3- 6x2 +12 x -8 - ( x3 - 27 ) + 6( x2 + 2x +1)
(=) x3 - 6x2 +12x -8 - x3 +27 + 6x2 +12x +6
(=) 24x + 25 = 49
(=) 24x = 49 - 25 = 24
(=) x = 24/24 =1
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
( x + 3)3 - x ( 3x+ 1)2 + ( 2x + 1) ( 4x2 - 2x + 1 ) = 28
<=> x3 + 9x2 + 27x + 27 - x ( 9x2 + 6x + 1) + 8x3+ 1 = 28
<=> x3 + 9x2 + 27x + 27 - 9x3 - 6x2 - x + 83 + 1
<=> 0 + 3x2 + 26x + 28 = 28
=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\3x=-26\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)
Vậy x1= 0 ; x2 = - 26/3
^^