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\({x^2} = {4^2} + {2^2} = 20 \Rightarrow x = 2\sqrt 5 \)
\({y^2} = {5^2} - {4^2} = 9 \Leftrightarrow y = 3\)
\({z^2} = {\left( {\sqrt 5 } \right)^2} + {\left( {2\sqrt 5 } \right)^2} = 25 \Rightarrow z = 5\)
\({t^2} = {1^2} + {2^2} = 5 \Rightarrow t = \sqrt 5 \)
CẢM ƠN NHÌU
1. 8 - 12x + 6x2 - x3
= 23 - 3.22.x + 3.x2.2 - x3
=(2-x)3
2. 125x3 - 75x2 +15x - 1
=(5x)3 - 3.(5x)2.1 + 3.5x.12 - 13
=(5x - 1)3
3, 4 (sai đề)
5. x3 + 2x2 - 6x - 27
=(x3 - 27) + (2x2 - 6x)
=(x3 - 33) + (2x2 - 6x)
=(x -3)(x2 + 3x + 9) + 2x(x-3)
=(x-3)(x2 + 3x +9 +2x)
=(x-3)(x2 + 5x +9)
6. 12x3 + 4x2- 27x -9
=(12x3 + 4x2) - (27x + 9)
=4x2(3x + 1) - 9(3x +1)
=(3x -1)(4x2 -9)
=(3x-1)(2x-3)(2x+3)
Bài 1:
\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^2\left(x^2-1\right)\)
\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
\(=x^6+27-27-27x^2-9x^4-x^6\)
\(=-9x^2\left(3-x^2\right)\)
Bài 5:
\(A=x^2-2x+1\)
\(=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
Vậy Min A = -2
Để A = -2 thì \(x-1=0\Rightarrow x=1\)
b, \(B=4x^2+4x+5\)
\(=\left(4x^2+4x+1\right)+4\)
\(=\left(2x+1\right)^2+4\)
Với mọi giá trị của x ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)
Vậy Min B = 4
Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
c, \(C=2x-x^2-4\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3
để C = -3 thì \(x-1=0\Rightarrow x=1\)
a, 4x.(x - 2017 ) - x + 2017 = 0
\(\Leftrightarrow\) 4x ( x - 2017 ) - ( x - 2017 ) = 0
\(\Leftrightarrow\) ( x - 2017 ) ( 4x - 1 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm x = 2017 hoặc x = \(\dfrac{1}{4}\) .
b) \(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-x-1\right)=0\)
\(x+1=0\)
x = -1
c) \(x\left(x-5\right)-\left(4x-20\right)=0\)
\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
a: \(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
b: Sửa đề: \(\left(x^3+6x^2+12x+8\right)+3\left(x^2+4x+4\right)+3\left(x+2\right)\)
\(=x^3+6x^2+12x+8+3x^2+12x+12+3x+6\)
\(=x^3+9x^2+27x+26\)
Bài 2:
a: 4x(x-3)+6(3-x)=0
=>4x(x-3)-6(x-3)=0
=>(x-3)(4x-6)=0
=>\(\left[{}\begin{matrix}x-3=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(x^3-x\left(x+1\right)\left(x-1\right)=14\)
=>\(x^3-x\left(x^2-1\right)=14\)
=>\(x^3-x^3+x=14\)
=>x=14
c: \(\left(x^2-x\right)^2+2\left(x^2-x\right)=8\)
=>\(\left(x^2-x\right)^2+2\left(x^2-x\right)-8=0\)
=>\(\left(x^2-x\right)^2+4\left(x^2-x\right)-2\left(x^2-x\right)-8=0\)
=>\(\left(x^2-x\right)\left(x^2-x+4\right)-2\left(x^2-x+4\right)=0\)
=>\(\left(x^2-x+4\right)\left(x^2-x-2\right)=0\)
=>\(\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}\right)\left(x-2\right)\left(x+1\right)=0\)
=>\(\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)