Tìm x, biết
\(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+1+6^x+1=7\cdot6^9\)
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Tìm x biết:
\(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^{x+1}+6^{x+1}=7\cdot6^9\)
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Tìm x:
\(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^{x+1}+6^{x+1}=7\cdot6^9\)
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\(\left(\frac{1}{2}-\frac{1}{3}\right).6^{x+1}+6^{x+1}=7.6^9\)
\(\Rightarrow\frac{1}{6}.6.6^x+6.6^x=7.6^9\)
\(\Rightarrow6^x+6.6^x=7.6^9\)
\(\Rightarrow6^x.\left(1+6\right)=7.6^9\)
\(\Rightarrow6^x=\frac{7.6^9}{7}=6^9\)
\(\Rightarrow x=9\)
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\(\left(\frac{1}{2}-\frac{1}{3}\right).6^{x+1}+6^{x+1}=7.6^9\)
\(\Leftrightarrow\frac{1}{6}.6^{x+1}+6^{x+1}=7.6^9\)
\(\Leftrightarrow6^{x+1}.\left(\frac{1}{6}+1\right)=7.6^9\)
\(\Leftrightarrow6^{x+1}.\frac{7}{6}=7.6^9\)
\(\Leftrightarrow6^{x+1}=7.6^9:\frac{7}{6}\)
\(\Leftrightarrow6^{x+1}=7.6^9.\frac{6}{7}\)
\(\Leftrightarrow6^{x+1}=\left(7.\frac{6}{7}\right).6^9\)
\(\Leftrightarrow6^{x+1}=6.6^9\)
\(\Leftrightarrow6^{x+1}=6^{10}\)
\(\Leftrightarrow x+1=10\)
\(\Leftrightarrow x=9\)
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Tìm x, biết
a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^7+6^4\)
b,\(7^{x+2}+2\cdot7^{x-1}=345\)
Bài 1: Tínha. left(1+frac{1}{1cdot3}right)cdotleft(1+frac{1}{2cdot4}right)cdotleft(1+frac{1}{3cdot5}right)+left(1+frac{1}{4cdot6}right).....left(1+frac{1}{99cdot101}right)b. left[sqrt{0,64}+sqrt{0,0001}-sqrt{left(-0,5right)^2}right]divleft[3cdotsqrt{left(0,04right)^2}-sqrt{left(-2right)^4}right]c. frac{5.4^{15}cdot9^9-4.3^{20}cdot8^9}{5cdot2^9cdot6^{19}-7cdot2^{29}cdot27^6}-frac{2^{19}cdot6^{15}-7cdot6^{10}cdot2^{20}cdot3^6}{9cdot6^{19}cdot2^9-4cdot3^{17}cdot2^{26}}+0,left(6right)Bài 2: Tìm x, y...
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Bài 1: Tính
a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)
b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)
c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)
Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)
b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\)
c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)
d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)
e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)
Bài 3: Chứng minh rằng
a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)
b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)
Bài 4:
a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)
b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)
c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
tìm x
\(\left(x+\frac{1}{2\cdot4}\right)+\left(x+\frac{1}{4\cdot6}\right)+\left(x+\frac{1}{6\cdot8}\right)+\left(x+\frac{1}{8\cdot10}\right)+\left(x+\frac{1}{10\cdot12}\right)=50\frac{5}{24}\)
\(\left(x\cdot6\frac{2}{7}+\frac{3}{7}\right)\cdot2\frac{1}{5}-\frac{3}{7}=-2\)
= ( x. 47/7 ) . 92/35 = -2 ( lưu ý (/ ) là dấu gạch ngang phân số )
=(x. 47 /7 )= 35/46
x=35/46 x 47 /7
x= 235/46
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Tìm số nguyên x, biết: \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2\cdot6}+\frac{2}{2.10}+.....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{2.6}+\frac{2}{2.10}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{x\left(x+1\right)}=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow2\left(1-\frac{1}{\left(x+1\right)}\right)=1\frac{1991}{1993}\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=1\frac{1991}{1993}\div2\)
\(\Leftrightarrow1-\frac{1}{\left(x+1\right)}=\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=1-\frac{1992}{1993}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)}=\frac{1}{1993}\)
\(\Leftrightarrow x=1992\)
\(\text{Vậy x = 1992 }\)
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tập hợp các số nguyên x thỏa mãn
\(x\cdot\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\right)<1\frac{6}{7}\)
Lời giải:
$x(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7})< 1\frac{6}{7}$
$x(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7})< \frac{13}{7}$
$x(1-\frac{1}{7})< \frac{13}{7}$
$x.\frac{6}{7}< \frac{13}{7}$
$x< \frac{13}{7}: \frac{6}{7}=\frac{13}{6}$
Vì $x$ là số nguyên nên $x\leq 2$
Vậy $x$ là các số nguyên sao cho $x\leq 2$.
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Bài 1:Tìm x
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2\cdot x+1\right)\cdot\left(2\cdot x+3\right)}=\frac{9}{19}\)
Bài 2: Tính nhanh
\(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2016\cdot2018}\)
ai giúp mình với gấp lắm không có bài là bị phạt đó
Bài 1 :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)
\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)
\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)
\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)
\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)
Bài 2 :
\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
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