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câu b:(x-1)(x+2)(x+3)(x+6)
= (x-1)(x+6)(x+2)(x+3)
= (x.x + 5.x - 6)(x.x + 5.x + 6)
đặt x.x + 5.x = t
=> (t -6)(t+6)
= t.t - 36
ta có:
t.t >= 0
suy ra t.t - 36 >= -36
vậy min = -36
dấu "=" xảy ra chỉ khi t.t = 0
chỉ khi x.x + 5.x = 0
chỉ khi x=0 hoặc x=-5
a) Ta có: A= 4x^2 + 4x + 11 = 4x^2 + 4x + 1 + 10
= (2x+1)^2 + 10 >= 10. A đạt giá trị nhỏ nhất = 10 khi x=-1/2
Mk lm câu c nhé, câu a và b bn tham khảo của ngô thế trường
\(c,C=x^2-2x+y^2-4y+7\)
\(C=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+2\)
\(C=\left(x-1\right)^2+\left(y-2\right)^2+2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\left(y-2\right)^2\ge0\forall y\)
\(2>0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\Rightarrow x=1\\\left(y-2\right)^2=0\Rightarrow y=2\end{cases}}\)
Vậy \(minC=2\Leftrightarrow x=1;y=2\)
hok tốt!
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
\(\frac{\left(x-2\right)^2}{12}-\frac{\left(x+1\right)^2}{21}=\frac{\left(x-4\right)\left(x-6\right)}{28}\)
<=> \(\frac{7\left(x^2-4x+4\right)}{84}-\frac{4\left(x^2+2x+1\right)}{84}=\frac{3\left(x^2-10x+24\right)}{84}\)
<=> 7x2 - 28x + 28 - 4x2 - 8x - 4 = 3x2 - 30x + 72
<=> 3x^2 - 36x - 3x^2 + 30x = 72 - 24
<=> -6x = 48
<=> x = -8
Vậy S = {-8}
e sẽ cố gắng !!!
\(3x-15=2x\left(x-5\right)\)
\(3x-15=2x^2-10x\)
\(3x-15-2x^2+10x=0\)
\(13x-15-2x^2=0\)
\(x\left(13-2x\right)-15=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(f,x\left(2x-7\right)-4x+14=0\)
\(2x^2-7x-4x+14=0\)
\(2x^2-11x+14=0\)
\(x\left(2x-11\right)=-14\)
\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)
1) Ta có: \(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2=0\)
\(\Leftrightarrow\left[\left(x^2-1\right)^2+x\left(x^2-1\right)\right]-\left[2x\left(x^2-1\right)+2x^2\right]=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+x-1\right)-2x\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left(x^2-2x-1\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-1=0\\x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=2\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\pm\sqrt{2}\\x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\pm\sqrt{2}\\x=-\frac{1\pm\sqrt{5}}{2}\end{cases}}\)
2) Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)
\(\Leftrightarrow\left[\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)\right]+\left[2x\left(x^2+4x+8\right)+2x^2\right]=0\)
\(\Leftrightarrow\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)=0\)
\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+5x+8\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)=0\)
Vì \(x^2+5x+8=\left(x^2+5x+\frac{25}{4}\right)+\frac{7}{4}=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)
Vậy x = -2 hoặc x = -4
\(\frac{x^2}{\left(x+y\right)\left(1-y\right)}-\frac{y^2}{\left(x+y\right)\left(1+x\right)}-\frac{x^2y^2}{\left(1+x\right)\left(1-y\right)}.\)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{x^2+x^3-y^2+y^3-x^3y^2-x^2y^3}{\left(x+y\right)\left(1-x\right)\left(1-y\right)}\)
\(=\frac{\left(x^2-y^2\right)+\left(x^3+y^3\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{\left(x+y\right)\left(x-y\right)+\left(x+y\right)\left(x^2-xy+y^2\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{\left(x+y\right)\left(x-y+x^2-xy+y^2-x^2y^2\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=\frac{2x+x^2+y^2-x^2y^2}{\left(1+x\right)\left(1-y\right)}\)
biến đổi thành nhân tử đi
khai triển ra thu gọn đc 2x3 + 10x = 12 => x = 1