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24 tháng 9 2021
\(a,2x^2+y^2+6x-2xy+9=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+6x+9\right)=0\\ \Leftrightarrow\left(x-y\right)^2+\left(x+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-3\end{matrix}\right.\Leftrightarrow x=y=-3\\ b,A=\left(x-2021\right)^2+\left(x+2022\right)^2=x^2-4042x+2021^2+x^2+4044x+2022^2\\ A=2x^2+2x+2021^2+2022^2\\ A=2\left(x^2+x+\dfrac{1}{4}\right)+2021^2+2022^2-\dfrac{1}{2}\\ A=2\left(x+\dfrac{1}{2}\right)^2+2021^2+2022^2-\dfrac{1}{2}\ge2021^2+2022^2-\dfrac{1}{2}\\ A_{max}=2021^2+2022^2-\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)\(c,P=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+16\\ P=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+16\\ P=\left(a^2+8a+11\right)^2-16+16=\left(a^2+8a+11\right)^2\left(Đpcm\right)\)
25 tháng 9 2021
1) \(\Leftrightarrow\left(x-4\right)\left(x+4\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4-x\right)=0\)
\(\Leftrightarrow\left(x-4\right)4=0\)
\(\Leftrightarrow x=4\)
2) \(\left(x+3\right)^2-\left(x-3\right)\left(x+5\right)=x^2+6x+9-x^2-2x+15=4x+24\)
3) \(2x^3+3x^2-2x+a=2x^2\left(x-2\right)+7x\left(x-2\right)+16\left(x-2\right)+32+a\)
Để \(2x^3+3x^2-2x+a⋮x-2\) thì \(32+a=0\Leftrightarrow a=-32\)
HP
25 tháng 9 2021
1.
x2 - 16 - x(x - 4) = 0
<=> (x2 - 42) - x(x - 4) = 0
<=> (x - 4)(x + 4) - x(x - 4) = 0
<=> (x + 4 - x)(x + 4) = 0
<=> 4(x + 4) = 0
<=> x + 4 = 0
<=> x = -4
2.
(x + 3)2 - (x - 3)(x + 5)
= x2 + 6x + 9 - (x2 + 5x - 3x - 15)
= x2 + 6x + 9 - x2 + 5x - 3x - 15
= x2 - x2 + 6x + 5x - 3x + 9 - 15
= 8x - 6
26 tháng 11 2021
\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
26 tháng 11 2021
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
23 tháng 5 2022
Câu 1:
a: x+2=0
nên x=-2
b: (x-3)(2x+8)=0
=>x-3=0 hoặc 2x+8=0
=>x=3 hoặc x=-4
DT
23 tháng 5 2022
a .
x + 2 = 0
=> x = 0 - 2 = -2
b ) .
<=> x - 3 = 0 ; 2x + 8 = 0
= > x = 3 ; x = -8/2 = -4
c ) .
ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
13 tháng 5 2022
a: A=(100-99)(100+99)+(98-97)(98+97)+...+(2-1)(2+1)
=100+99+98+...+2+1
=5050
b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)+1
\(=2^{64}-1+1=2^{64}\)
20 tháng 2 2022
a: \(A=\left(100-99\right)\left(100+99\right)+\left(98+97\right)\left(98-97\right)+....+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1\)
=5050
b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\cdot\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1=2^{128}\)
20 tháng 2 2022
a. \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=199+195+...+3\)
\(=\dfrac{\left(199+3\right)\left(\dfrac{199-3}{4}+1\right)}{2}=5050\)
b. \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-2b^2-4ab\)
\(=2c^2\)
ta có \(a\left(a+3\right)\left(a+1\right)\left(a+2\right)+1=0\Leftrightarrow\left(a^2+3a\right)\left(a^2+3a+2\right)+1=0\)
đặt \(a^2+3a+1=x\)
=> ta có \(\left(x-1\right)\left(x+1\right)+1=0\Leftrightarrow x^2=0\Leftrightarrow x=0\Leftrightarrow a^2+3a+1=0\) vô nghiệm => a không tồn tại
câu 2 tương tự
mình cung làm ra thế rồi nhưng nghi có thể ra hơn chứ ko vô nghiệm