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\(b,=1^2-\left(x-y\right)^2=\left(1+x-y\right)\left(1-x+y\right)\)
\(c,=\left(x^2+1\right)^2-\left(2x\right)^2=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)^2\left(x-1\right)^2\)
giúp mình vớiiiii ạ
Bài 4
50 mm = 5 cm
Thể tích hình hộp chữ nhật:
V = AB . BC . AA' = 3 . 4 . 5 = 60 (cm³)
Gọi S(km) là quãng đường để 2 xe gặp nhau(t>0)
Đổi: \(15ph=\dfrac{1}{4}h\)
Theo đề bài ta có: \(\left\{{}\begin{matrix}S=t_1.v_1=15t_1\\S=t_2.v_2=\left(t_1-\dfrac{1}{4}\right).45=-\dfrac{45}{4}+45t_1\end{matrix}\right.\)
\(\Rightarrow15t_1=-\dfrac{45}{4}+45t_1\Rightarrow t_1=\dfrac{3}{8}\left(h\right)\)
Cách điểm a: \(S=t_1.v_1=\dfrac{3}{8}.15=5,625\left(km\right)\)
\(=\left(28^2-27^2\right)+\left(26^2-25^2\right)+...+\left(2^2-1^2\right)\))
\(=\left(28+27\right)\left(28-27\right)+\left(26-25\right)\left(26+25\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=28+27+26+25+...+2+1\)
= 28 x 29 / 2 = 406
(a^2+b^2+c^2)^2-(a^2+b^2-c^2)
\(=a^2+b^2+c^2+2ab+2bc+2ac-a^2-b^2+c^2\)
\(=2c^2+2ab+2bc+2ac\)
\(=2\left(c^2+ab+bc+ac\right)=2\left[\left(c^2+ac\right)+\left(ab+bc\right)\right]\)
\(=2\left[c\left(a+c\right)+b\left(a+c\right)\right]=2\left(a+c\right)\left(b+c\right)\)
k nha!
\(\frac{2}{x-2}-\frac{3}{x+2}=\frac{x+1}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{x^2-4}=0\)
\(\Leftrightarrow\frac{2}{x-2}-\frac{3}{x+2}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{3\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x+4-3x+6-x-1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-2x-9}{\left(x-2\right)\left(x+2\right)}=0\)
=> -2x-9=0
<=> -2x=9
<=> \(x=\frac{-9}{2}\left(tmđk\right)\)
Ta có: \(1\cdot99^2+2\cdot98^2+\cdots+99\cdot1^2\)
\(=99^2\left(100-99\right)+98^2\left(100-98\right)+\cdots+1^2\left(100-1\right)\)
\(=100\left(1^2+2^2+\cdots+99^2\right)-\left(1^3+2^3+\cdots+99^3\right)\)
\(=100\cdot\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}-\left(1+2+\cdots+99\right)^2\)
\(=100\cdot\frac{99\cdot100\cdot199}{6}-\left\lbrack99\cdot\frac{100}{2}\right\rbrack^2\)
\(=100\cdot33\cdot50\cdot199-\left(99\cdot50\right)^2\)
\(=5000\cdot33\cdot199-99^2\cdot50^2=8332500\)