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a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
a, \(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}\)
\(=\left(\frac{1}{4}+\frac{5}{12}\right)-\left(\frac{1}{13}+\frac{7}{8}\right)\)
\(=\frac{2}{3}-\frac{99}{104}\)
\(=-\frac{89}{312}\)
b, \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)
\(=\left(11\frac{3}{13}+5\frac{3}{13}\right)-2\frac{4}{7}\)
\(=\frac{214}{13}-\frac{18}{7}\)
\(=\frac{1264}{91}\)
c, \(\left(6\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)
\(=6\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)
\(=\left(6\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)
\(=2+3\frac{7}{11}\)
\(=5\frac{7}{11}\)
\(=\frac{62}{11}\)
d, \(\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-0,25-\frac{1}{12}\right)\)
\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\cdot0\)
\(=0\)
e, \(-1,5\cdot\left(1+\frac{2}{3}\right)\)
\(=-\frac{3}{2}\cdot\frac{5}{3}\)
\(=-\frac{5}{2}\)
f, Đặt \(A=1^2+2^2+3^2+...+100^2\)
\(=1+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)
\(=1+2\cdot3-2+3\cdot4-3+...+100\cdot101-100\)
\(=\left(2\cdot3+3\cdot4+...+100\cdot101\right)-\left(1+2+3+...+100\right)\)
Đặt B = 2 . 3 + 3 . 4 + ... + 100 . 101
3B = 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 100 . 101 . ( 102 - 99 )
3B = 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 100 . 101 . 102 - 99 . 100 . 101
3B = 100 . 101 . 102
B = \(\frac{100\cdot101\cdot102}{3}\)
B = 343400
Thay B vào A. Ta được :
\(A=343400-\left(1+2+3+...+100\right)\)
Thay C = 1 + 2 + 3 + ... + 100
Dãy số 1; 2; 3; ...; 100 có số số hạng là:
( 100 - 1 ) : 1 + 1 = 100 ( số hạng )
Tổng của dãy số đó là :
( 100 + 1 ) . 100 : 2 = 5050
=> C = 5050
Thay C vào A. Ta được :
\(A=343400-5050\)
\(A=338350\)
Vậy A = 338350
a,-200 x10 t10z3
b,\(\frac{-5}{4}\)x11 y5 z4
c,\(\frac{2}{15}\)x6 y6 z9
d,\(\frac{1}{7}\)x10 y6 z7
e,-4z6 y10 z6
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
a)\(23\frac{1}{4}\cdot\frac{7}{5}+13\frac{1}{4}\)/\(\frac{-5}{7}\)
=\(23\frac{1}{4}\cdot\frac{7}{5}+13\frac{1}{4}\cdot\frac{-7}{5}\)
=\(23\frac{1}{4}\cdot\frac{7}{5}+\left(-13\frac{1}{4}\right)\cdot\frac{7}{5}\)
=\(\frac{7}{5}\cdot\left(23\frac{1}{4}-13\frac{1}{4}\right)\)
=\(\frac{7}{5}\cdot\left(23-13+\frac{1}{4}-\frac{1}{4}\right)\)
=\(\frac{7}{5}\cdot10\)
=\(14\)
b)\(\left(-3\right)^2\cdot\left(\frac{3}{4}-0.25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
=\(9\cdot\left(\frac{3}{4}-\frac{1}{4}\right)-\left(3-1+\frac{1}{2}-\frac{1}{2}\right)\)
=\(9\cdot\frac{1}{2}-2\)
=\(\frac{9}{2}-2\)
=\(\frac{9}{2}-\frac{4}{2}\)
=\(\frac{5}{2}\)
Nếu mình còn sai hay thiếu chỗ nào thì ae bảo mình nha :))