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Bài 2:
a: \(A=\left[a+\left(b-c\right)\right]^2+\left[a-\left(b-c\right)\right]^2\)
\(=a^2+2a\left(b-c\right)+\left(b-c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2\)
\(=2a^2+2\left(b-c\right)^2\)
\(=2\cdot1^2+2\left(3+1\right)^2=2+32=34\)
b: \(B=a^2+2ab+b^2-a^2+2ab-b^2=4ab=4\cdot2\cdot5=40\)
a: A=(2x-1)^3
Khi x=5,5 thì A=(2*5,5-1)^3=10^3=1000
b: B=27x^3+54x^2+36x+7
=(3x)^3+3*(3x)^2*2+3*3x*2^2+2^3-1
=(3x+2)^3-1
=(-8+2)^3-1
=(-6)^3-1=-217
a) (2x - 5)2 - (5 + 2x) = 0
<=> 4x2 - 22x + 20 = 0
\(\Leftrightarrow\left(2x-\dfrac{11}{2}\right)^2=\dfrac{41}{4}\)
\(\Leftrightarrow x=\dfrac{\pm\sqrt{41}+11}{4}\)
b) \(27x^3-54x^2+36x=0\)
\(\Leftrightarrow x\left(3x^2-6x+4\right)=0\)
\(\Leftrightarrow x=0\) (Vì \(3x^2-6x+4=3\left(x-1\right)^2+1>0\forall x\))
c) x3 + 8 - (x + 2).(x - 4) = 0
\(\Leftrightarrow\left(x+2\right).\left(x^2-2x+4\right)-\left(x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-3x+8\right)=0\)
\(\Leftrightarrow x=-2\) (Vì \(x^2-3x+8=\left(x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\))
d) \(x^6-1=0\)
\(\Leftrightarrow\left(x^2\right)^3-1=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^4+x^2+1\right)=0\)
\(\Leftrightarrow x^2-1=0\) (Vì \(x^4+x^2+1>0\))
\(\Leftrightarrow x=\pm1\)
\(d,x^6-1=0\\ \Leftrightarrow\left(x^2\right)^3-1^3=0\\ \Leftrightarrow\left(x^2-1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x^4+x^2+1=0\left(Vô.lí,vì:x^4\ge0;x^2\ge0,\forall x\in R\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ c,\left(x^3+8\right)-\left(x+2\right)\left(x-4\right)=0\\ \Leftrightarrow\left(x^3+8\right)-\left(x^2-2x-8\right)=0\\ \Leftrightarrow x^3-x^2+2x+16=0\\ \Leftrightarrow x^3+2x^2-3x^2-6x+8x+16=0\\ \Leftrightarrow x^2\left(x+2\right)-3x\left(x+2\right)+8\left(x+2\right)=0\\ \Leftrightarrow\left(x^2-3x+8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-3x+8=0\left(Vô.lí\right)\\x+2=0\end{matrix}\right.\Leftrightarrow x=-2\)
a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)
\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)
\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
b) \(27x^3-54x^2+36x=9\)
\(\Rightarrow27x^3-54x^2+36x-9=0\)
\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)
\(\Rightarrow\left(3x-2\right)^3-1=0\)
\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)
mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)
\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)
(\(x-5\))2 = (3 +2\(x\))2 ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}
27\(x^3\) - 54\(x^2\) + 36\(x\) = 9
27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1
(3\(x\) - 2)3 = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
a, (x+3)2 - ( 2x + 1 ).( x+3)=0 b, x3-12x2+36x =0
=> (x+3).(x+3-2x-1) => x(x2-12x+36) = 0
=>(x+3).(-x+2) => x(x-6)2 = 0
=> x+3=0 <=> x=-3 => x=0 <=> x=0
-x+2=0 <=> x=-2 x-6= 0 <=> x=6
Bài 5
a) A = -x³ + 6x² - 12x + 8
= -x³ + 3.(-x)².2 - 3.x.2² + 2³
= (-x + 2)³
= (2 - x)³
Thay x = -28 vào A ta được:
A = [2 - (-28)]³
= 30³
= 27000
b) B = 8x³ + 12x² + 6x + 1
= (2x)³ + 3.(2x)².1 + 3.2x.1² + 1³
= (2x + 1)³
Thay x = 1/2 vào B ta được:
B = (2.1/2 + 1)³
= 2³
= 8
Bài 6
a) 11³ - 1 = 11³ - 1³
= (11 - 1)(11² + 11.1 + 1²)
= 10.(121 + 11 + 1)
= 10.133
= 1330
b) Đặt B = x³ - y³ = (x - y)(x² + xy + y²)
= (x - y)(x² - 2xy + y² + 3xy)
= (x - y)[(x - y)² + 3xy]
Thay x - y = 6 và xy = 9 vào B ta được:
B = 6.(6² + 3.9)
= 6.(36 + 27)
= 6.63
= 378
a) \(A=-x^3+6x^2-12x+8\)
\(A=-\left(x^3-6x^2+12x-8\right)\)
\(A=-\left(x-2\right)^3\)
Thay x=-28 vào A ta có:
\(A=-\left(-28-2\right)^3=27000\)
Vậy: ...
b) \(B=8x^3+12x^2+6x+1\)
\(B=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3\)
\(B=\left(2x+1\right)^3\)
Thay \(x=\dfrac{1}{2}\) vào B ta có:
\(B=\left(2\cdot\dfrac{1}{2}+1\right)^3=8\)
Vậy: ...
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: \(B=\left(a+b-a+b\right)\left(a+b+a-b\right)=4ab=4\cdot2\cdot5=40\)