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a)
Áp dụng công thức (a - b).(a+ b) = a.(a+ b) - b.(a+ b) = a2 + ab - ab - b2 = a2 - b2
Ta có
\(M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
M = (100 - 99)(100 + 99) + (98 - 97).(98 + 97) + ...+ (2 - 1)(2+1)
= 100 + 99 + 98 + 97 + ...+ 2 + 1
= (1+100).100 : 2
= 5050
b)
N = (202 - 192 ) + (182 - 172 ) + ...+ (42 - 32 ) + (22 - 12 )
= (20 - 19).(20 + 19) + (18 - 17)(18 + 17) +...+ (4 -3)(4 +3) + (2-1)(2+1) = 39 + 35 + ...+ 7 + 3
N = (39 + 3).10 : 2 = 210
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
\(1,a,\left(-\frac{3}{4}\right)^0=1\)
\(b,\left(-2\frac{1}{3}\right)^4=\left[-\left(\frac{2\cdot3+1}{3}\right)\right]^4=\left(\frac{-7}{3}\right)^4=\frac{2401}{256}\)
\(c,\left(2,5\right)^3=15,625\)
\(d,25^3:5^2=5^6:5^2=5^4=625\)
\(e,2^2\cdot4^3=2^2\cdot2^6=2^8=256\)
\(f,\left(\frac{1}{5}\right)^5\cdot5^3=\left(\frac{1}{5}\right)^5:\frac{1}{5^3}=\left(\frac{1}{5}\right)^5:\left(\frac{1}{5}\right)^3=\left(\frac{1}{5}\right)^2=\frac{1}{25}\)
\(g,\left(\frac{1}{5}\right)^3\cdot10^3=\left(\frac{1}{5}\cdot10\right)^3=2^3=8\)
\(h,\left(-\frac{2}{3}\right)^4:2^4=\left(-\frac{2}{3}:2\right)^4=\left(-\frac{1}{3}\right)^4=\frac{1}{81}\)
\(i,\left(\frac{2}{3}\right)^4:9^2=\left(\frac{2}{3}\right)^4:3^4=\left(\frac{2}{3}:3\right)^4=\left(\frac{2}{9}\right)^4=\frac{16}{6561}\)
\(k,\left(\frac{1}{2}\right)^3.\left(\frac{1}{4}^2\right)=\left(\frac{1}{2}\right)^3.\left(\frac{1}{2}\right)^4=\left(\frac{1}{2}\right)^7=\frac{1}{128}\)
\(m,\left(\frac{120}{40}\right)^3=3^3=27\)
\(n,\frac{390^4}{130^4}=\left(\frac{390}{130}\right)^4=3^4=81\)
\(2,a,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}\cdot\frac{1}{2}\)
\(=2+\frac{1}{8}\)
\(=\frac{17}{8}\)
\(b,\left(-2\right)^3+2^2+\left(-1\right)^{20}+\left(-2\right)^0\)
\(=-8+4+1+1\)
\(=-2\)
\(c,\left(3^2\right)^2-\left(\left(-5\right)^2\right)^2+\left(\left(-2\right)^3\right)^2\)
\(=3^4-\left(-5\right)^4+\left(-2\right)^6\)
\(=81-625+64\)
\(=-480\)
\(3,a,\left(x-1\right)^3=27\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
\(b,x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\Rightarrow x=-1\end{cases}}\)
\(c,\left(2x+1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=4\\2x=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
\(d,\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2=1\Rightarrow x-1=\pm1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\text{ or }x=0\end{cases}}\)
\(4,M=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(M=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(M=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(M=100+99+98+97+...+2+1\)
\(M=\left(100+1\right)\cdot100:2\)
\(M=101\cdot50=5050\)
Bài 3:
a: \(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy=x^2+11xy-y^2\)
b: \(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2=-x^2+10xy-12y^2\)
Bài 2:
\(A+B=4x^4-5xy+5y^2+3x^2+2xy-y=4x^4+3x^2-3xy+5y^2-y\)
\(A-B=4x^4-5xy+5y^2-3x^2-2xy+y=4x^4-3x^2+5y^2-7xy+y\)
\(B-A=-\left(A-B\right)=-4x^4+3x^2-5y^2+7xy-y\)
1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)