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1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(a,\dfrac{-3}{4}x+1=\dfrac{5}{6}\\ \Rightarrow\dfrac{-3}{4}x=\dfrac{-1}{6}\\ \Rightarrow x=\dfrac{2}{9}\\ b,\left(2x-3\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\\ c,\dfrac{1}{2}-\left|x+1\right|=0,25\\ \Rightarrow\left|x+1\right|=0,25\\ \Rightarrow\left[{}\begin{matrix}x+1=0,25\\x+1=-0,25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-0,75\\x=-1,25\end{matrix}\right.\)
a: =>-3/4x=-1/6
hay x=2/9
b: =>2x-3=0 hoặc x-5=0
hay x=3/2 hoặc x=5
c: =>|x+1|=1/4
\(\Leftrightarrow x+1\in\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
hay \(x\in\left\{-\dfrac{3}{4};-\dfrac{5}{4}\right\}\)
Hoàng Ngọc Anh bài này nè bn giúp mk nha!!! ngày mai mk phải nộp bài rùi =.=
a) \(\Rightarrow\dfrac{\dfrac{7}{2}x+\dfrac{59}{24}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Rightarrow\left(\dfrac{7}{2}x+\dfrac{59}{24}\right).52=\dfrac{13}{30}.137\)
\(\Rightarrow182x+\dfrac{767}{6}=\dfrac{1781}{30}\)
\(\Rightarrow x=\dfrac{-79}{210}\)
b) Tương tự câu a)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=2\\x+\dfrac{1}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\\x=-\dfrac{11}{5}\end{matrix}\right.\)
\(a,\left|x+\dfrac{1}{5}\right|-4=-2\\ \left|x+\dfrac{1}{5}\right|=\left(-2\right)+4\\ \left|x+\dfrac{1}{5}\right|=2\Rightarrow x+\dfrac{1}{5}=2\)
hoặc \(x+\dfrac{1}{5}=-2\)
Với \(x+\dfrac{1}{5}=2\Rightarrow x=2-\dfrac{1}{5}\)hay \(x=\dfrac{9}{5}\)
Với \(x+\dfrac{1}{5}=-2\Rightarrow x=-2-\dfrac{1}{5}\)hay \(x=-\dfrac{11}{5}\)
Thông cảm . bt làm mỗi câu a
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=-\dfrac{2}{6}\)
\(\Rightarrow x=-\dfrac{2}{6}-\dfrac{1}{3}\)
\(\Rightarrow x=-\dfrac{2}{3}\)
b) \(\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
\(\Rightarrow\left(\dfrac{15}{40}-\dfrac{8}{40}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{7}{40}+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{5}{8}-x=\dfrac{1}{5}-\dfrac{7}{40}\)
\(\Rightarrow\dfrac{5}{8}-x=\dfrac{1}{40}\)
\(\Rightarrow x=\dfrac{5}{8}-\dfrac{1}{40}\)
\(\Rightarrow x=\dfrac{3}{5}\)
Sửa câu a:
(x - 2)2 - 36 = 0
(x - 2 - 6)(x - 2 + 6) = 0
(x - 8)(x + 4)= 0
\(\Leftrightarrow \begin{bmatrix} x - 8= 0 & & \\ x + 4 = 0 & & \end{bmatrix}\)
\(\Leftrightarrow \begin{bmatrix} x = 8 & & \\ x = - 4 & & \end{bmatrix}\)
pn bỏ dấu ngoặc bên phải nhé
Vậy x = 8; x = - 4
2:
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{2a}{2b}=\dfrac{a}{b}\)
Từ đó suy ra \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\RightarrowĐPCM\)