Tìm x: \(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
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24 tháng 3 2019
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
\(\Leftrightarrow x+116=0\Leftrightarrow x=-116\)
24 tháng 3 2019
\(\frac{x-1}{117}+\frac{x-2}{118}+\frac{x-3}{119}=\frac{x-4}{120}+\frac{x-5}{121}+\frac{x-6}{122}\)
\(\Leftrightarrow\frac{x-1}{117}+1+\frac{x-2}{118}+1+\frac{x-3}{119}+1=\frac{x-4}{120}+1+\frac{x-5}{121}+1+\frac{x-6}{122}+1\)
\(\Leftrightarrow\frac{x+116}{117}+\frac{x+116}{118}+\frac{x+116}{119}-\frac{x+116}{120}-\frac{x+116}{121}-\frac{x+116}{122}=0\)
\(\Leftrightarrow\left(x+116\right)\left(\frac{1}{117}+\frac{1}{118}+\frac{1}{119}-\frac{1}{120}-\frac{1}{121}-\frac{1}{122}\right)=0\)
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
SN
5 tháng 7 2016
Bài 1:
Thay \(x=\frac{4}{3};y=-1\)vào biểu thức A, ta được:
\(A=\frac{4}{3}\cdot\left[3\cdot\frac{4}{3}-\left(-1\right)\right]-\left(3\cdot\frac{4}{3}+1\right)\left(-1\right)\)
\(A=\frac{20}{3}+5=\frac{35}{3}\)
Vậy khi \(x=\frac{4}{3};y=-1\)thì A=\(\frac{35}{3}\)
\(B=3\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{8}{39}\)
\(B=\frac{352}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot\frac{713}{119}-\frac{8}{39}=-\frac{412}{1071}\)
27 tháng 1 2017
a, Đặt \(x=\frac{1}{117}\), \(y=\frac{1}{119}\) ta có:
\(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+24x\)
\(=3y+xy-24x+4xy-5xy+24x\)
\(=3y\)
\(=\frac{3}{119}\)
b, Thay 8 bằng x + 1 ta có:\(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x^2+x-5\)
\(=7-5\)
= 2
BT
2
2 tháng 12 2017
Đặt \(x^2+1=a\)
\(\Rightarrow\frac{a}{120}+\frac{a+1}{119}+\frac{a+2}{118}=3\)
\(\Leftrightarrow21241a=2506200\)
\(\Leftrightarrow a=\frac{2506200}{21241}\)
\(\Rightarrow x=.....\)
2 tháng 12 2017
\(\frac{x^2}{120}+\frac{x^2+1}{119}+\frac{x^2+2}{118}=3\)
\(\Leftrightarrow\frac{x^2}{120}+1+\frac{x^2+1}{119}+1+\frac{x^2+2}{118}+1=6\)
\(\Leftrightarrow\frac{x^2+120}{120}+\frac{x^2+120}{119}+\frac{x^2+120}{118}=6\)
\(\Leftrightarrow\left(x^2+120\right)\left(\frac{1}{120}+\frac{1}{119}+\frac{1}{118}\right)=6\)
\(\Leftrightarrow x^2+120=\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}}\)
\(\Leftrightarrow x^2=\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}}-1\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}}-1}\\x=-\sqrt{\frac{6}{\frac{1}{120}+\frac{1}{119}+\frac{1}{118}-1}}\end{cases}}\)
5 tháng 9 2018
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
5 tháng 9 2018
Gợi ý:
Đặt:
\(\frac{1}{117}=a\)
\(\frac{1}{119}=b\)
Đến đây bạn thế a, b vào A rồi thu gọn, sau đó tính
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
8 tháng 12 2018
Đặt biểu thức trong ngoặc là A
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{7.8.9.10}.\)
\(3A=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+\frac{6-3}{3.4.5.6}+...+\frac{10-7}{7.8.9.10}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{7.8.9}-\frac{1}{8.9.10}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{8.9.10}\Rightarrow A=\frac{1}{1.2.3.3}-\frac{1}{3.8.9.10}\)
Từ đó tính ra x . Bạn tự làm nốt nhé. Ngại tính
AN
2
25 tháng 5 2018
\(\frac{x+1}{125}+\frac{x+2}{124}+\frac{x+3}{123}+\frac{x+4}{122}+\frac{x+146}{5}=0\)
\(\left(\frac{x+1}{125}+1\right)+\left(\frac{x+2}{124}+1\right)+\left(\frac{x+3}{123}+1\right)+\left(\frac{x+4}{122}+1\right)+\left(\frac{x+146}{5}-4\right)=0\)
\(\frac{x+126}{125}+\frac{x+126}{124}+\frac{x+126}{123}+\frac{x+126}{122}+\frac{x+126}{5}=0\)
\(\left(x+126\right).\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)=0\)
vì \(\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)\ne0\)nên x + 126 = 0 \(\Rightarrow\)x = -126