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a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
\(\dfrac{-1}{4}x+\dfrac{2}{3}=\dfrac{5}{9}\)\(\Rightarrow\)\(\dfrac{-1}{4}x=\dfrac{5}{9}-\dfrac{2}{3}\)=\(\dfrac{-1}{9}\)
\(\Rightarrow\) x = \(\dfrac{-1}{9}:\dfrac{-1}{4}\)=\(\dfrac{4}{9}\).
\(x.\left(\dfrac{3}{5}\right)^3=\dfrac{3}{5}\)
\(\Rightarrow\)x=\(\dfrac{3}{5}:\left(\dfrac{3}{5}\right)^3=\left(\dfrac{3}{5}\right)^{-2}\)= \(2\dfrac{7}{9}\)
\(\left|x\right|\) + \(\dfrac{1}{5}=2-\left(\dfrac{2}{3}-\dfrac{3}{4}\right)\)=2 - \(\dfrac{-1}{12}\)=2\(\dfrac{1}{12}\)
\(\Rightarrow\)\(\left|x\right|\)=\(2\dfrac{1}{12}\)-\(\dfrac{1}{5}\)=\(1\dfrac{53}{60}\)
\(\Rightarrow\)x=\(\left[{}\begin{matrix}1\dfrac{53}{60}\\-1\dfrac{53}{60}\end{matrix}\right.\)
\(\left(\dfrac{-3}{4}\right)^x=\dfrac{81}{256}\)=\(\dfrac{(-3)^4}{4^4}\)=\(\left(\dfrac{-3}{4}\right)^4\)
\(\Rightarrow\) x = 4
14 Mẹo Vặt Sáng Tạo Dành Cho Học Sinh - YouTube
thử xem mẹo thứ 5 đi chứ theo mk đây là bài toán dễ lớp 6
Bài 5:
a: \(\dfrac{x+7}{x-5}< 0\)
=>x+7>0 và x-5<0
=>-7<x<5
b: \(\dfrac{x-4}{x+9}>0\)
=>x-4>0 hoặc x+9<0
=>x>4 hoặc x<-9
c: \(\dfrac{x-1}{x+9}>1\)
\(\Leftrightarrow\dfrac{x-1-x-9}{x+9}>0\)
=>x+9>0
hay x>-9
d: \(\dfrac{x+5}{x-11}< 1\)
\(\Leftrightarrow\dfrac{x+5-x+11}{x-11}< 0\)
=>x-11<0
hay x<11
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
Tìm x dễ thì tự làm nha:
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)-\left(\dfrac{x+2}{2002}+1\right)-\left(\dfrac{x+1}{2003}\right)=0\)\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
1/a/ \(\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\\dfrac{2}{5}-3x=-\dfrac{3}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
b/ \(\left(\dfrac{2}{3}x-\dfrac{1}{5}\right)^5=\dfrac{1}{243}\)
\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{5}=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{8}{15}\)
\(\Leftrightarrow x=\dfrac{4}{5}\)
Vậy .........
2/ a/
Ta có :
\(5^{222}=\left(5^2\right)^{111}=25^{111}\)
\(2^{555}=\left(2^5\right)^{111}=32^{111}\)
Vì \(25^{111}< 32^{111}\Leftrightarrow5^{222}< 2^{555}\)
b/ Ta có :
\(3^{48}=\left(3^4\right)^{12}=81^{12}\)
\(4^{36}=\left(4^3\right)^{12}=64^{12}\)
Vì \(81^{12}>64^{12}\Leftrightarrow3^{48}>4^{36}\)
Cái này dễ lắm. Mình giải luôn nhé!
a) \(\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\Leftrightarrow x=\dfrac{2}{7}:\dfrac{1}{7}\Leftrightarrow x=2\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\Leftrightarrow x=-\dfrac{3}{5}:\left(-\dfrac{1}{5}\right)\Leftrightarrow x=3\\\dfrac{1}{3}x+\dfrac{4}{3}=0\Leftrightarrow x=-\dfrac{4}{3}:\dfrac{1}{3}\Leftrightarrow x=-4\end{matrix}\right.\)
Vậy x=2 hoặc x=3 hoặc x=-4
b)\(x\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)+1=0\)
\(x.0+1=0\)
\(1=0\) ( vô lí)
Vậy không có giá trị của x nào thỏa mãn
Cái này bạn áp dụng tính chất 1 của tỉ lệ thức là ra ngay mà!
Hai tỉ số bằng nhau khi tích 2 ngoại tỉ bằng tích 2 trung tỉ.
nhìn đã nản
e) \(\dfrac{x-5}{7}=\dfrac{x+2}{3}\Leftrightarrow3\left(x-5\right)=7\left(x+2\right)\Leftrightarrow3x-15=7x+14\)
\(\Leftrightarrow7x-3x=-15-14\Leftrightarrow4x=-29\Leftrightarrow x=\dfrac{-29}{4}\) vậy \(x=\dfrac{-29}{4}\)
f) \(\left(x-5\right)\left(x+2\right)=\left(x-3\right)\left(x+4\right)\Leftrightarrow x^2+2x-5x-10=x^2+4x-3x-12\)
\(\Leftrightarrow x^2+2x-5x-10-x^2-4x+3x+12=0\Leftrightarrow-4x+2=0\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\)
vậy \(x=\dfrac{1}{2}\)