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Câu c) Sử dụng hằng đẳng thức+Đặt biến phụ
Ta có: \(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
\(=\left(x+y\right)\left(x+y-1\right)-12\)
Đặt: \(x+y=t\)
\(=t\left(t-1\right)-12\)
\(=t^2-t-12\)
\(=t^2-t-9-3\)
\(=\left(t^2-3^2\right)-\left(t+3\right)\)
\(=\left(t+3\right)\left(t-3\right)-\left(t+3\right)\)
\(=\left(t+3\right)\left(t-4\right)\)Bn tự thế vào nhá. (Bài c) tương tự bài a))
Câu d) Đặt biến phụ
Ta có: \(\left(5x^2-2x\right)^2+2x-5x^2-6\)
\(=\left(5x^2-2x\right)^2-5x^2+2x-6\)
\(=\left(5x^2-2x\right)^2-\left(5x^2-2x\right)-6\)
\(=\left(5x^2-2x\right)\left(5x^2-2x-1\right)-6\)
Đặt \(t=5x^2-2x\)
\(=t\left(t-1\right)-6\)
\(=t^2-t-6\)
\(=t^2-t-9+3\)
\(=\left(t^2-3^2\right)-\left(t-3\right)\)
\(=\left(t-3\right)\left(t+3\right)-\left(t-3\right)\)
\(=\left(t-3\right)\left(t+2\right)\)Bn tự thế t vào
Câu a) Sử dụng phương pháp đặt biến phụ+hằng đẳng thức
Ta có: \(\left(2x^2+x-2\right)\left(2x^2+x-3\right)-12\)
Đặt: \(t=2x^2+x-2\)
\(=t\left(t-1\right)-12\)
\(=t^2-t-12=t^2-t-9-3\)
\(=\left(t^2-3^2\right)-\left(t+3\right)\)
\(\left(t+3\right)\left(t-3\right)-\left(t+3\right)=\left(t+3\right)\left(t-4\right)\)
Thay t vào: \(\left(2x^2+x+1\right)\left(2x^2+x-6\right)\)
Câu b) Sử dụng hằng đẳng thức+ đặt biến phụ
Ta có: \(x^2+9y^2-9y-3x+6xy+2\)
\(=\left(x^2+6xy+9y^2\right)-\left(9y+3x\right)+2\)
\(=\left(x+3y\right)^2-3\left(3y+x\right)+2\)
\(=\left(x+3y\right)\left(x+3y-3\right)+2\)
Đặt \(t=x+3y\)
\(=t\left(t-3\right)+2\)
\(=t^2-3t+2\)
\(=\left(t^2-4\right)-\left(3t-6\right)\)
\(=\left(t-2\right)\left(t+2\right)-3\left(t-2\right)\)
\(=\left(t-2\right)\left(t-1\right)\)Khúc sau bn tự thế vào
Còn mấy bài sau đang nghiên cứu
\(5x^2+5y^2+8xy-2x+2y+2=0\)
\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+1\right)^2\ge0\)
\(\Rightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(\left(x+y\right)^{2018}+\left(x-2\right)^{2019}+\left(y+1\right)^{2020}=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}=-1\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
2x2 + 2y2 + 3xy - x + y + 1 = 0
2x2 + 2y2 + 4xy - xy - x + y + 1 = 0
(2x2 + 2y2 + 4xy) + (-xy - x) + (y + 1) = 0
2(x + y)2 - x(y + 1) + (y + 1) = 0
2(x + y)2 + (y + 1)(1 - x) = 0
Do (x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0
\(\Rightarrow y+1=0;1-x=0\)
*) y + 1 = 0
y = -1
*) 1 - x = 0
x = 1
Với x = 1; y = -1, ta có:
B = [1 + (-1)]2018 + (1 - 2)2018 + (-1 - 1)2018
= 1 + 22018
Đặt \(\left\{{}\begin{matrix}2018-x=a\\x-2019=b\end{matrix}\right.\) \(\Rightarrow a+b=-1\Rightarrow b=-1-a\)
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5a=-3b\\3a=-5b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5a=-3\left(-1-a\right)\\3a=-5\left(-1-a\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=3\\2a=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2018-x=\frac{3}{2}\\2018-x=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{4033}{2}\\x=\frac{4041}{2}\end{matrix}\right.\)
À khác cái dấu nhưng đề phải là giải phương trình chứ
Đặt 2017-x=a => x-2018=-a-1 phương trình trở thành:
\(\frac{a^2+a\left(-a-1\right)+\left(a-1\right)^2}{a^2-a\left(-a-1\right)+\left(a-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{3}{2}\\a=-\frac{5}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=2015,5\\x=2019,5\end{cases}}}\)
Vậy......................
BPT\(\Leftrightarrow\left(2017x^2+2018\right)\left(2x-1\right)-\left(2017x^2+2018\right)\left(4-5x\right)\ge0\)
\(\Leftrightarrow\left(2017x^2+2018\right)\left(2x-1-4+5x\right)\ge0\)
\(\Leftrightarrow\left(2017x^2+2018\right)\left(7x-5\right)\ge0\)
DO 2017x2+2018 luôn luôn lớn hơn 0
ĐỂ B PT \(\ge\)0\(\Leftrightarrow7x-5\ge0\)
\(\Leftrightarrow x\ge\frac{5}{7}\)
vậy ...........