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Câu 3:
a: \(\left(x+2\right)^2=x^2+4x+4\)
b: \(\left(x+3\right)^2=x^2+6x+9\)
c: \(\left(x-3\right)^2=x^2-6x+9\)
d: \(\left(x-7\right)^2=x^2-14x+49\)
e: \(x^2-6x+9=\left(x-3\right)^2\)
f: \(x^2-8x+16=\left(x-4\right)^2\)
g: \(=\left(x-10\right)\left(x+10\right)\)
h: \(=\left(x-11\right)\left(x+11\right)\)
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\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)
\(=-0,2\)
\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(=x^3-8y^3-x^3+8y^3-10\)
\(=-10\)
\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)
\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=13\)
a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)
\(A=-\dfrac{1}{5}\)
Vậy: ...
b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)
\(B=-10\)
Vậy: ...
c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)
\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)
\(=13\)
Vậy:...
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Bài 2:
a: =>168x+20=6x-21
=>162x=-41
hay x=-41/162
b: \(\Leftrightarrow2\left(3x-8\right)=3\left(5-x\right)\)
=>6x-16=15-3x
=>9x=31
hay x=31/9
c: \(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x+4\right)\left(x+10\right)=3\left(x^2+2x-8\right)\)
\(\Leftrightarrow4x^2+32x-80-x^2-14x-40-3x^2-6x+24=0\)
=>12x-96=0
hay x=8
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\(-2\left(2x-7\right)^2=2\)
\(\Rightarrow\left(2x-7\right)^2=-4\)
Mà: \(\left(2x-7\right)^2\ge0\)
=> Ko có giá trị x cần tìm
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\(2,\\ a,=2x^2+4x-3x-6-2x^2-4x-2=-3x-8\\ b,=\left[x-2+2\left(x+1\right)\right]^2=\left(x-2+2x+2\right)^2=9x^2\)
20.
a) \(A=4x^2-4x+m\)
\(=4x^2-4x+1-1+m\)
\(=\left(4x^2+4x+1\right)+\left(m-1\right)\)
\(=\left(2x+1\right)^2+\left(m-1\right)\)
Để biếu thức không âm thì \(A\ge0\Rightarrow\left(2x+1\right)^2+\left(m-1\right)\ge0\)
Mà: \(\left(2x+1\right)^2\ge0\)
\(\Rightarrow\) Để \(A\ge0\Rightarrow m-1\ge0\Leftrightarrow m\ge1\)
b) \(B=x^2-6x+2-m\)
\(=x^2-6x+9-9+2-m\)
\(=\left(x^2+6x+9\right)+\left(2-m-9\right)\)
\(=\left(x+3\right)^2+\left(-m-7\right)\)
Để bt không âm thì \(B\ge0\Rightarrow\left(x+3\right)^2+\left(-m-7\right)\ge0\)
Mà: \(\left(x+3\right)^2\ge0\)
⇒ Để \(B\ge0\Rightarrow-m-7\ge0\Leftrightarrow-m\ge7\Leftrightarrow m\le-7\)
Câu 16:
a) Số chính phương lẻ có dạng: \(\left(2x+1\right)^2=4x^2+4x+1=4x\left(x+1\right)+1\)
Vì \(x\in N\Rightarrow x\left(x+1\right)\) là tích của 2 số tự nhiên liên tiếp
\(\Rightarrow x\left(x+1\right)\) ⋮ 2
\(\Rightarrow4x\left(x+1\right)\) ⋮ 8
\(\Rightarrow4x\left(x+1\right)+1\) chia 8 dư 1
b) Theo câu a ta biết số chính phương lẻ chia 8 sẽ dư 1 mà `1^2;3^2;5^2;7^2;...;2023^2` đều là các số tự nhiên lẻ ⇒ Chúng đều chia 8 dư 1
Từ 1 đến 2023 có số lượng số lẻ là: \(\left(2023-1\right):2+1=1012\) (số)
Khi đó `1^2+3^2+5^2+...+2023^2` chia 8 dư `1*1012=1012`
Mà: 1012 chia 8 dư 4 ⇒ Tổng `1^2+3^2+5^2+...+2023^2` chia 8 dư 4