![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 5:
a: Ta có: \(A=\left(x-1\right)\left(x-3\right)+11\)
\(=x^2-4x+3+11\)
\(=x^2-4x+4+10\)
\(=\left(x-2\right)^2+10\ge10\forall x\)
Dấu '=' xảy ra khi x=2
b: Ta có: \(B=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
Câu 5:
a) \(A=\left(x-1\right)\left(x-3\right)+11=x^2-4x+3+11\)
\(=x^2-4x+14\)
\(=\left(x^2-4x+4\right)+10=\left(x-2\right)^2+10\ge10\)
\(minA=10\Leftrightarrow x=2\)
b) \(B=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: Ta có: \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Leftrightarrow x^2-8x+16-x^2+4=6\)
\(\Leftrightarrow-8x=-14\)
hay \(x=\dfrac{7}{4}\)
c: Ta có: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow5x^2+2x+10-5x^2+245=0\)
\(\Leftrightarrow x=-\dfrac{255}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bn chụp màn hình sao vậy? k bt giống bn k alt+ctrl+a
Bài 5:
a: \(x\left(x-1\right)-x^2+4x=-3\)
\(\Leftrightarrow x^2-x-x^2+4x=-3\)
hay x=-1
b: \(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=7\)
\(\Leftrightarrow-11x=-3\)
hay \(x=\dfrac{3}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{A}{x-3}=\dfrac{y-x}{3-x}\left(Đk:x\ne3\right)\)
\(A=\dfrac{\left(x-3\right)\left(y-x\right)}{3-x}=x-y\)
b) \(\dfrac{5x}{x+1}=\dfrac{Ax\left(x-1\right)}{\left(1-x\right)\left(x+1\right)}\left(Đk:x\ne\pm1\right)\)
\(A=\dfrac{5x\left(1-x\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=-5\)
c) \(\dfrac{4x^2-5x+1}{A}=\dfrac{4x-1}{x+3}\left(Đk:x\ne-3;A\ne0\right)\)
\(A=\dfrac{\left(4x^2-5x+1\right)\left(x+3\right)}{4x-1}=\dfrac{\left(x-1\right)\left(4x-1\right)\left(x+3\right)}{4x-1}\)
\(=\left(x-1\right)\left(x+3\right)=x^2+2x-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
Ta có: \(\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
\(=6x^2+9x+14x+21-\left(6x^2+33x-10x-55\right)\)
\(=6x^2+23x+21-6x^2-23x+55\)
=76
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
\(A=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ B=x^3-y^3-5+2y^3-x^3-y^3=-5\\ C=x^3-3x^2+3x-1-x^3-3x^2-3x-1-6x^2+6=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3) \(\sqrt{\left(x-2\right)\left(x+1\right)}\) thì (x-2)(x+1)>0
=> x2 -x-2>0
=> x2 - x - \(\dfrac{1}{2}\)- \(\dfrac{3}{2}\)>0
= (x+\(\dfrac{1}{4}\))2 - 3/2 >0
=> x+ 1/4>3/2
=> x>5/4
4) Có x đâu mà tìm bạn??