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a) x2 + 6x + 9 = (x + 3)2
b) 25x2 - 30x + 9 = (5x - 3)2
c) (2x - 3y)(2x + 3y) = (2x)2 - (3y)2 = 4x2 - 9y2
d) (x + 2y)(x2 + 2xy + 4y2) = (x + 2y)3
e) 27x3 + 54x2 + 36x + 8
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
Bài 2:
a: \(A=a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2-2ab-2bc+2ac\)
\(=2a^2+2b^2+2c^2-4bc\)
\(=2+2\cdot9+2\cdot1-4\cdot3\cdot\left(-1\right)=22+12=34\)
b: \(B=\left(a+b-a+b\right)\left(a+b+a-b\right)=4ab=4\cdot2\cdot5=40\)
c) 5(x^2+8x+16)+4(x^2-10x+25)-9(x^2-16)
=5x^2+40x+80+4x^2-40x+100-9x^2+144
=80+100+144
=324
a, x2-x+1/4=(x-1/2)2
b, (x+1)3
c,(2x+1)3
d, (2-3x03
e, (10x)2-(x2+25)2=:[10x+(x2+25)][10x-(x2+25)]=(10x+x2+25)(10x-x2-25)
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
a. x.(x+3)-x2+15=0
=> x^2 + 3x - x^2 + 15 = 0
=> 3x + 15 = 0
=> 3x = -15
=> x = -5
vậy_
b. (2x-1)(x+3) - x(2x-6) =15
=> 2x^2 + 6x - x - 3 - 2x^2 + 6x = 15
=> x - 3 = 15
=> x = 18
vậy_
c. x3 -36x = 0
=> x(x^2 - 36) = 0
=> x = 0 hoặc x^2 - 36 = 0
=> x = 0 hoặc x^2 = 36
=> x = 0 hoặc x = 6 hoặc x = -6
vậy_
d. 6x2 + 6x =x2+2x+1
=> 6x(x + 1) = (x + 1)^2
=> 6x(x + 1) - (x + 1)^2 = 0
=> (x + 1)(6x - x - 1) = 0
=> (x + 1)(5x - 1) = 0
=> x = -1 hoặc 5x = 1
=> x = -1 hoặc x = 1/5
vậy_
e. x(3x+1)=1-9x2
=> x(3x + 1) = (1 - 3x)(1 + 3x)
=> x(3x + 1) - (1 - 3x)(1 + 3x) = 0
=> (3x + 1)(x - 1 + 3x) = 0
=> (3x + 1)(4x - 1) = 0
=> 3x + 1 = 0 hoặc 4x - 1 = 0
=> 3x = -1 hoặc 4x = 1
=> x = -1/3 hoặc x = 1/4
vậy_
a: \(27x^3-54x^2+36x-8=0\)
=>\(\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2+3\cdot3x\cdot2^2-2^3=0\)
=>\(\left(3x-2\right)^3=0\)
=>3x-2=0
=>3x=2
=>\(x=\frac23\)
b: \(x^3+15x^2+75x+125=0\)
=>\(x^3+3\cdot x^2\cdot5+3\cdot x\cdot5^2+5^3=0\)
=>\(\left(x+5\right)^3=0\)
=>x+5=0
=>x=-5
c: \(x^3-18x^2+108x-216=0\)
=>\(x^3-3\cdot x^2\cdot6+3\cdot x\cdot6^2-6^3=0\)
=>\(\left(x-6\right)^3=0\)
=>x-6=0
=>x=6
d: \(\left(x-1\right)^3-x\left(x^2+3x\right)=2\)
=>\(x^3-3x^2+3x-1-x^3-3x^2=2\)
=>\(-6x^2+3x-3=0\)
=>\(2x^2-x+1=0\)
\(\Rightarrow x^2-\frac12x+\frac12=0\)
=>\(x^2-\frac12x+\frac{1}{16}+\frac{7}{16}=0\)
=>\(\left(x-\frac14\right)^2+\frac{7}{16}=0\) (vô lý)
=>x∈∅
e: \(\left(x+1\right)^3+\left(x-2\right)^3-2x^2\left(x-1,5\right)=3\)
=>\(x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=3\)
=>15x-7=3
=>15x=10
=>\(x=\frac{10}{15}=\frac23\)