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a/ \(\left|-x\right|=1,5\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)
Vậy .....
b/ \(\left|x+\dfrac{1}{2}\right|=2\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{5}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy ....
c/ \(\left|0,5-x\right|=\left|-0,5\right|\)
\(\left|0,5-x\right|=0,5\)
\(\Leftrightarrow\left[{}\begin{matrix}0,5-x=0,5\\0,5-x=-0,5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy ...
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=t\) \(\Rightarrow a=bt\);\(c=dt\)
rồi bạn thế vào điều phải chứng minh là ra
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
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
a/ \(\left|-1,3\right|-\left|-3,7\right|+\left|-\dfrac{1}{2}\right|\)
\(=1,3-3,7+\dfrac{1}{2}\)
\(=-2,4+\dfrac{1}{2}\)
\(=-2,9\)
b/ \(\left|\dfrac{2}{5}\right|-\left|-0,2\right|.\left|-7\right|\)
\(=\dfrac{2}{5}-0,2.7\)
\(=\dfrac{2}{5}.1,4\)
\(=0,56\)
c/ \(-15:\left|-3\right|+\left|0,5\right|\)
\(=-15:2+0,5\)
\(=-7,5+0,5\)
\(=-8\)
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.
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
\(\dfrac{-1}{4}x+\dfrac{2}{3}=\dfrac{5}{9}\)\(\Rightarrow\)\(\dfrac{-1}{4}x=\dfrac{5}{9}-\dfrac{2}{3}\)=\(\dfrac{-1}{9}\)
\(\Rightarrow\) 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
Ta có:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
\(\Rightarrow\dfrac{7a-11b}{7c-11d}=\dfrac{4a+5b}{4c+5d}\)
\(\Leftrightarrow\dfrac{7a}{7c}=\dfrac{11b}{11d}=\dfrac{4a}{4c}=\dfrac{5b}{5d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Mặt khác:
\(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)= k
Vì \(\dfrac{a}{b}=k\) = > a = bk
Vì \(\dfrac{c}{d}=k\) = > c = dk
Ta có: \(\dfrac{7a-11b}{4a+5b}=\dfrac{7.bk-11b}{4.bk+5b}=\dfrac{\left(7.11\right).b.\left(k-1\right)}{\left(4.5\right).b.\left(k+1\right)}\dfrac{\left(7.11\right).\left(k-1\right)}{\left(4.5\right).\left(k+1\right)}\)(1)
\(\dfrac{7c-11d}{4c+5d}=\dfrac{7.dk-11d}{4.dk+5d}=\dfrac{\left(7.11\right).d.\left(k-1\right)}{\left(4.5\right).d.\left(k+1\right)}=\dfrac{\left(7.11\right).\left(k-1\right)}{\left(4.5\right).\left(k+1\right)}\left(2\right)\)Từ (1) và (2) = > \(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
a/ \(x+\dfrac{3}{5}=\dfrac{4}{7}\)
\(x=\dfrac{4}{7}-\dfrac{3}{5}\)
\(x=-\dfrac{1}{35}\)
Vậy ....
b/ \(x-\dfrac{5}{6}=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+\dfrac{5}{6}\)
\(x=1\)
Vậy ....
c/\(-\dfrac{5}{7}-x=\dfrac{-9}{10}\)
\(x=\dfrac{-5}{7}-\dfrac{-9}{10}\)
\(x=\dfrac{13}{70}\)
Vậy .....
d/ \(\dfrac{5}{7}-x=10\)
\(x=\dfrac{5}{7}-10\)
\(x=\dfrac{-65}{7}\)
Vậy ...
e/ \(x:\left(\dfrac{1}{9}-\dfrac{2}{5}\right)=\dfrac{-1}{2}\)
\(x:\dfrac{-13}{45}=\dfrac{-1}{2}\)
\(x=\dfrac{-1}{2}.\dfrac{-13}{45}\)
\(x=\dfrac{13}{90}\)
Vậy ....
f/ \(\left(\dfrac{-3}{5}+1,25\right)x=\dfrac{1}{3}\)
\(0,65.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:0,65\)
\(x=\dfrac{20}{39}\)
Vậy ....
g/ \(\dfrac{1}{3}x+\left(\dfrac{2}{3}-\dfrac{4}{9}\right)=\dfrac{-3}{4}\)
\(\dfrac{1}{3}x+\dfrac{2}{9}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{-35}{36}\)
\(\Leftrightarrow x=\dfrac{-35}{12}\)
Vậy ...