Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)(2x-3)2=(x+5)2
=>4x2-12x+9=x2+10x+25
=>3x2-22x-16=0
=>3x2+2x-24x-16=0
=>x(3x+2)-8(3x+2)=0
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
b)X2.(x-1)-4x2+8x-4=0
=>x2(x-1)-4x2+4x+4x-4=0
=>x2(x-1)-4x(x-1)-4(x-1)=0
=>x2(x-1)-(4x-4)(x-1)=0
=>(x2-4x+4)(x-1)=0
=>(x-2)2(x-1)=0
=>x=2 hoặc x=1
c) 4x2- 25 - (2x- 5) . ( 2x+7)=0
=>4x2-25-(4x2+14x-10x-35)=0
=>4x2-25-4x2-14x+10x+35=0
=>-4x+10=0
=>-4x=-10 <=>x=5/2
d) x3+27+(x+3).(x-9)=0
=>x3+33+(x+3)(x-9)=0
=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
=>(x2-3x+9+x-9)(x+3)=0
=>(x2-2x)(x+3)=0
=>x(x-2)(x+3)=0
=>x=0 hoặc x=2 hoặc x=-3
e) (x-2).(x+5)- x2+4=0
=>(x-2)(x+5)-(x-2)(x+2)=0
=>(x-2)(x+5-x-2)=0
=>3(x-2)=0 <=>x=2
Sau khi khai triển hằng đẳng thức và thực hiện chuyển vế bạn sẽ đk kết quả như này!(\(\left(2x-3\right)^2=\left(x+5\right)^2=3x^2-22x-14\)
a) \(x^2+4x+4=\left(x+2\right)^2\)
b) \(x^2-8x+16=\left(x-4\right)^2\)
c) \(\left(x+5\right)\left(x-5\right)=x^2-25\)
d) \(x^2+2x+1=\left(x+1\right)^2\)
e) \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)
f) \(\left(2x+3y\right)^2+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)
g) \(\left(2+bx^2\right)\left(bx^2-2\right)=\left(bx^2+2\right)\left(bx^2-2\right)=\left(bx^2\right)^2-4=b^2x^4-4\)
bài 1 điền vào chỗ trống
a) x2 + 4x + 4
= (x + 2)2
b) x2 - 8x + 16
= (x - 4)2
c) x3 +12x2 + 48x + 64
= (x + 4)3
d) x3 - 6x + 12x - 8
= (x - 2)3
e) x2 + 2x + 1
= (x + 1)2
f) x2 - 1
= (x - 1)(x + 1)
g) x2 - 4x + 4
= (x - 2)2
h) x2 - 4
= (x - 2)(x + 2)
i) x2 + 6x + 9
= (x + 3)2
j) 4x2 - 9
= (2x - 3)(2x + 3)
k) 16x2 - 8x + 1
= (4x - 1)2
l) 9x2 + 6x + 1
= (3x + 1)2
m) 36x2 + 36x + 9
= (6x + 3)2
n) x3 + 27
= (x + 3)(x2 - 3x + 9)
o) 17x3 + 27 (Đề sai)
a) \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
b) \(16x^2-8x+1=\left(4x\right)^2-2.4x.1+1^2=\left(4x-1\right)^2\)
c) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)
d) \(36x^2+36x+9=\left(6x\right)^2+2.6x.3+3^2=\left(6x+3\right)^2\)
e) \(x^3+27=x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)
a) (2x+3)(4x2-6x+9)-2(4x3-1)+(8x-1)=15
<=>8x3+27-8x3+2+8x-1=15
<=>8x+28=15
<=>8x=-13
<=>x=-13/8
b) (x+3)3-(x+9)(x2+27)-(5x-216) = 3x-4
<=>x3+9x2+27x+27-x3-27x-9x2-243-5x+216=3x-4
<=>-5x=3x-4
<=>8x=4
<=>x=1/2
a ) \(\left(x+2\right)^3-\left(x-2\right)^3\)
\(=\left[\left(x+2\right)-\left(x-2\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right]\)
\(a.\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-5\right)=71\)
\(\Leftrightarrow x^3+1-x^3+5x=71\)
\(\Leftrightarrow5x=71-1\)
\(\Leftrightarrow5x=70\)
\(\Leftrightarrow x=70:5=14\)
\(b.\left(2x-3\right)^3-8x\left(x-1\right)^2+4x\left(4x+1\right)+27=0\)
\(\Leftrightarrow8x^3-12x^2+18x-27-8x\left(x^2-2x+1\right)+16x^2+4x+27=0\)
\(\Leftrightarrow8x^3-12x^2+18x-27-8x^3+16x^2-8x+16x^2+4x+27=0\)
\(\Leftrightarrow20x^2+14x=0\)
\(\Leftrightarrow x\left(20x+14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\20x+14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{7}{10}\end{cases}}}\)
a) ta có: (x+1)(x^2 -x+1) -x(x^2 -5)=71
<=>x^3 +1 -x^3 +5x=71
<=>5x=70
<=>x=14
b) ta có:(2x-3)^3 -8x(x-1)^2 +4x(4x+1)+27=0
<=>[ (2x-3)^3 +27)] - [ 8x(x-1)^2 -4x(4x+1)]=0
<=> (2x-3+3)[ (2x-3)^2 - (2x-3).3 +3^2] - 2x [ 4(x^2 -2x +1) -2(4x+1)]=0
<=>2x( 4.x^2 - 12x +9 - 6x +9 +9) - 2x( 4.x^2 -8x+4 -8x -2)=0
<=>2x(4.x^2 -18x +27) - 2x(4.x^2 -16x +2)=0
<=>2x(4.x^2 -18x+27 -4.x^2 +16x-2)=0
<=>2x(25-2x)=0
<=>x=0 hoặc 25-2x=0 <=> x=0 hoặc x=25/2
a) \(x^2+4x+4=\left(x+2\right)^2\)
b) \(x^2-8x+16=\left(x-4\right)^2\)
c) \(\left(x+5\right)\left(x-5\right)=x^2-25\)
g) \(\left(x-2\right)\left(x^2-2x+4\right)\)
\(=x^3-8\)