Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 1:
a. \((x+1)(x+3) - x(x+2)=7 \)
\(x^2+ 3x +x +3 - x^2 -2x =7\)
\(x^2+4x+3-x^2-2x=7\)
\(=> 2x+3=7\)
\(2x=4\)
\(x = 2\)
Bài 2:
a)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
Bài 1:
a: \(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
b: Để A>0 thì 1-2x>0
=>2x<1
=>x<1/2
Bài 2:
\(A=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1\)
\(A_{max}=-1\) khi \(x=2\)
\(B=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(B_{max}=7\) khi \(x=2\)
\(C=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(C_{max}=\frac{1}{4}\) khi \(x=\frac{1}{2}\)
\(D=-\left(x^2-2x+1\right)-\left(y^2-4y+4\right)+11\)
\(D=-\left(x-1\right)^2-\left(y-2\right)^2+11\le11\)
\(D_{max}=11\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
\(E=-\frac{1}{2}\left(4x^2-4x+1\right)-\frac{9}{2}=-\frac{1}{2}\left(2x-1\right)^2-\frac{9}{2}\le-\frac{9}{2}\)
\(E_{max}=-\frac{9}{2}\) khi \(x=\frac{1}{2}\)
Bài 1:
\(A=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1\)
\(A_{min}=1\) khi \(x+1=0\Leftrightarrow x=-1\)
\(B=\left(x-3\right)^2\ge0\)
\(B_{min}=0\) khi \(x=3\)
\(C=2\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2+\frac{9}{2}\ge\frac{9}{2}\)
\(C_{min}=\frac{9}{2}\) khi \(x=\frac{3}{2}\)
\(D=\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)
\(D=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(D_{min}=\frac{3}{4}\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=-3\end{matrix}\right.\)
Bài 2 :
Câu a : \(y\left(y^3+y^2-y-2\right)-\left(y^2-2\right)\left(y^2+y+1\right)\)
\(=y^4+y^3-y^2-2y-y^4-y^3-y^2+2y^2+2y+2\)
\(=2\) \(\Rightarrow\) ko phụ thuộc vào biến .
Câu b : \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)
\(=29\Rightarrow\) ko thuộc vào biến
Câu c : \(3x\left(x+5\right)-\left(3x+18\right)\left(x-1\right)\)
\(=3x^2+15x-3x^2+3x-18x+18\)
\(=18\) \(\Rightarrow\) ko thuộc vào biến
Câu d : \(\left(2x+6\right)\left(4x^2-12x+36\right)-8x^3+5\)
\(=8x^3-24x^2+72x+24x^2-72x+216-8x^3+5\)
\(=221\) \(\Rightarrow\) không thuộc vào biến
câu 1) a) \(\left(x^2+2xy+y^2\right)\left(x+y\right)=\left(x+y\right)^2\left(x+y\right)=\left(x+y\right)^3\)
b) \(y\left(y^3+y^2-3y-2\right)+\left(y^2-2\right)\left(y^2+y-1\right)\)
\(=y^4+y^3-3y^2-2y+y^4+y^3-y^2-2y^2-2y+2\)
\(=2y^4+2y^3-6y^2-4y+2=2y\left(y^3+y^2-3y-2\right)+2\)
\(=2y\left(y+2\right)\left(y^2-y-1\right)+2=2\left(y^2+2y\right)\left(y^2-y-1\right)+2\)
\(=2\left(y^2+2y\right)\left(y^2-y-1+1\right)=2\left(y^2+2y\right)\left(y^2-y\right)\)
c) \(6x^2-\left(2x+5\right)\left(3x-2\right)=6x^2-\left(6x^2-4x+15x-10\right)\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-11x+10\)
d) \(\left(2x-1\right)\left(3x+1\right)+\left(3x+4\right)\left(3-2x\right)\)
\(\)\(=6x^2+2x-3x-1+9x-6x^2+12-8x=11\)
e) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)\)
\(=21x-15x^2-35+25x-\left(10x-15x^2+4-6x\right)\)
\(21x-15x^2-35+25x-10x+15x^2-4+6x=42x-39\)
Bài 2:
\(A=x^2+4y^2-2x+10-4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 vào biểu thức A ta được: \(A=5^2-2.5+10=25\)
\(B=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)\left(y-1\right)+y^2-2y+1\)
\(=x^2+4xy+4y^2-2xy+2x-4y^2+4y+y^2-2y+1\)
\(=x^2+2xy+y^2+2x+2y+1\)
\(=\left(x+y\right)^2+2\left(x+y\right)+1\)
Thay x + y = 5 vào biểu thức B ta được: \(B=5^2+2.5+1=25+10+1=36\)
\(C=x^2-y^2-4x=\left(x^2-4x+4\right)-y^2-4\)
\(=\left(x-2\right)^2-y^2-4\) \(=\left(x-y-2\right)\left(x-2+y\right)-4\)
Thay x + y = 2 vào C ta được: \(C=\left(x-2-y\right)\left(2-2\right)-4=0-4=-4\)
\(D=x^2+y^2+2xy-4x-4y-3\)
\(=\left(x+y\right)^2-4\left(x+y\right)-3\) Thay x + y = 4 vào D ta được:
\(D=4^2-4.4-3=16-16-3=-3\)
Bài 3:
a) \(N=-9x^2+12x-5=-\left(9x^2-12x+4\right)-1\)
\(=-\left(3x-2\right)^2-1\)
Do \(\left(3x-2\right)^2\ge0\) nên \(-\left(3x-2\right)^2-1< 0\)
Vậy N < 0
b) ghi đề cẩn thận lại đi, mk k hiểu
1.
a. Đặt x-7 = t
\(\Rightarrow x-3=t+4;x-11=t-4\)
\(\Rightarrow\left(x-3\right)^2+\left(x-11\right)^2=\left(t+4\right)^2+\left(t-4\right)^2=t^2+16+8t+t^2+16-8t=2t^2+32\)
Vì \(2t^2\ge0\) nên: \(2t^2+32\ge32\)
Dấu "=" xảy ra \(\Leftrightarrow2t^2=0\)
\(\Leftrightarrow t^2=0\)
\(\Leftrightarrow\left(x-7\right)^2=0\)
\(\Leftrightarrow x-7=0\Leftrightarrow x=7\)
Vậy \(Min_A=32\Leftrightarrow x=7\)