B=1+2+4+5+7+8+10+...+119+121+122
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c: C=1*2+2*3+3*4+...+58*59+59*60
=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)
=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60
=>3*C=59*60*61
=>C=59*20*61=71980
A= -1 - (2-3-4+5) - (6+7+8-9) -... - (198 -199-120+121)
A= -1 - 0-0-...-0
A= -1
nhớ k giùm mk
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)
\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
Vì \(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\ne0\)
Nên x + 116 = 0
<=> x = -116
chỉ cần giải cho mình câu C bài 1 và câu B,C bài 2 thôi nhé
Bài 1:
a) A = 210+211+212
=210*(1+21+22)
=210*(1+2+4)
=7*210 chia hết 7
Đpcm
b)7*32=244
=32+64+128
=25+26+27
\(B=1+2+4+5+7+8+10+...+119+121+122\)
Ta có :
\(A=1+2+3+...+121+122\)
\(A=\left[\left(122-1\right):1+1\right]\left(1+122\right):2\)
\(A=122.\left(123\right):2=7503\)
Ta lại có :
\(C=3+6+9+...+120\)
\(C=\left[\left(120-3\right):3+1\right]\left(3+120\right):2\)
\(C=40.123:2=2460\)
Ta thấy : \(B=A-C\)
\(B=7503-2460=5043\)