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Lời giải:
Ta có:
$(2x-y)+(2y-z)+(2z-x)=1+2+3$
$2x-y+2y-z+2z-x=6$
$x+y+z=6$
1: Ta có: \(S_1=1+\left(-2\right)+3+\left(-4\right)+...+\left(-2020\right)+2021\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(2019-2020\right)+2021\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2021\)
\(=-1\cdot1010+2021\)
\(=-1010+2021=1011\)
2) Ta có: \(S_2=\left(-2\right)+4+\left(-6\right)+8+...+\left(-2014\right)+2016\)
\(=\left(-2+4\right)+\left(-6+8\right)+...+\left(-2014+2016\right)\)
\(=2+2+...+2\)
\(=2\cdot504=1008\)
Ta có: A = 1.2.3 + 2.3.4 + 3.4.5 + ... + 18.19.20
=> 4A = 4(1.2.3 + 2.3.4 + 3.4.5 + ... + 18.19.20)
=> 4A = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 +...... +18.19.20.4
=> 4A = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6 - 2) +..... +18.19.20.(21 - 17)
=> 4A = 1.2.3.4 + 2.3.4.5 - 2.3.4 + 3.4.5.6 - 2.3.4.5 + ..... + 17.18.19.20 - 17.18.19.20
=> 4A =18.19.20
=> 4A = 6840
=> A = 1710
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+18.19.20
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+18.19.20)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+18.19.20.(21-17)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+ 18.19.20.21-17.18.19.20
=>A =18.19.20.21
=>A+18.19.20.21/4
=> A=35910
\(\left(x+3\right)+\left(x+4\right)+\left(x+5\right)+...+\left(x+22\right)=450\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(3+4+5+...+22\right)=450\) ( 20 số x )
\(\Rightarrow20x+250=450\)
\(\Rightarrow20x=200\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\) => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) => \(\frac{a}{b}=\frac{13}{6}\)
b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)
a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)
<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)
<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)
<=>\(\frac{a}{b}=\frac{13}{6}\)
b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3
=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3
=\(\frac{23}{4}\).\(\frac{5}{3}\)-3
=\(\frac{115}{12}\)-3
=\(\frac{115-36}{12}\)
=\(\frac{79}{12}\)
\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{3}{4}+\frac{3}{4}\right)-2=?\)
what?