\(\frac{-1}{4}-\frac{3}{4}\div x=\frac{-11}{36}\)
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Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\frac{x-5}{3}=\frac{y-4}{4}=\frac{z-3}{5}=\frac{x-5+y-4+z-3}{3+4+5}=2\)
\(\Rightarrow x-5=2.3=6\Rightarrow x=11\); \(y-4=2.4=8\Rightarrow y=12\); \(z-3=2.5=10\Rightarrow z=13\)
\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
\(x^3+3x^2+3x+1+y^3+3y^3+3y+1+x+y+2=0\)
\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right)+\left(x+y+2\right)=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)+1\right)=0\)
\(\Leftrightarrow x+y+2=0\)
(phần trong ngoặc \(\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\frac{\left(y+1\right)^2}{4}+\frac{3\left(y+1\right)^2}{4}+1\)
\(=\left(x+1-\frac{y+1}{4}\right)^2+\frac{3\left(y+1\right)^2}{4}+1\) luôn dương)
\(\Rightarrow x+y=-2\)
Mà \(xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-x>0\\-y>0\end{matrix}\right.\)
Ta có: \(\frac{1}{-x}+\frac{1}{-y}\ge\frac{4}{-\left(x+y\right)}=2\) \(\Leftrightarrow\frac{1}{x}+\frac{1}{y}\le-2\) (đpcm)
Dấu "=" xảy ra khi và chỉ khi \(x=y=-1\)
2/ \(x;y;z\ne0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{xz+yz+z^2}\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(\frac{xy+yz+xz+z^2}{xyz\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\) dù trường hợp nào thì thay vào ta đều có \(B=0\)
3/ \(\Leftrightarrow mx-2x+my-y-1=0\)
\(\Leftrightarrow m\left(x+y\right)-\left(2x+y+1\right)=0\)
Gọi \(A\left(x_0;y_0\right)\) là điểm cố định mà d đi qua
\(\Leftrightarrow\left\{{}\begin{matrix}x_0+y_0=0\\2x_0+y_0+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_0=-1\\y_0=1\end{matrix}\right.\)
Vậy d luôn đi qua \(A\left(-1;1\right)\) với mọi m
a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)
\(\frac{11}{4}:\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{4}:\frac{7}{2}\)
\(\Leftrightarrow\frac{3}{2}:\left|4x-\frac{1}{3}\right|=\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{3}{2}:\frac{11}{14}\)
\(\Leftrightarrow\left|4x-\frac{1}{3}\right|=\frac{21}{11}\)
\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=\frac{21}{11}\\4x-\frac{1}{3}=-\frac{21}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{37}{66}\\x=-\frac{13}{33}\end{cases}}\)
Bài dưới tương tự
\(\Rightarrow\frac{3}{4}:x=-\frac{1}{4}-\left(-\frac{11}{36}\right)=\frac{1}{18}\Rightarrow x=\frac{3}{4}:\frac{1}{18}=\frac{27}{2}\)
\(\frac{3}{4}:x=\frac{-1}{4}+\frac{11}{36}\)
\(\frac{3}{4}:x=\frac{-9}{36}+\frac{11}{36}\)
\(\frac{3}{4}:x=\frac{1}{18}\)
\(x=\frac{3}{4}:\frac{1}{18}=\frac{3}{4}.\frac{18}{1}=\frac{27}{2}\)
\(\left(4\frac{1}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{14}\)
\(\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{14}.\frac{7}{4}=\frac{11}{8}\)
\(\frac{2}{5}x=\frac{9}{2}-\frac{11}{8}\)
\(\frac{2}{5}x=\frac{25}{8}\)
\(x=\frac{25}{8}.\frac{5}{2}=\frac{125}{16}\)
\(\left(4\frac{1}{2}-\frac{2}{5}x\right):\frac{7}{4}=\frac{11}{14}\)
\(\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{14}x\frac{7}{4}\)
\(\left(\frac{9}{2}-\frac{2}{5}x\right)=\frac{11}{8}\)
\(\frac{2}{5}x=\frac{9}{12}-\frac{11}{8}\)
\(\frac{2}{5}x=\frac{-5}{8}\)
\(x=\frac{-5}{8}:\frac{2}{5}\)
\(x=\frac{-25}{16}\)
\(\frac{6\div\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}=\frac{6\div\frac{3}{5}-\frac{1\times6+1}{6}\times\frac{6}{7}}{\frac{4\times5+1}{5}\times\frac{10}{11}+\frac{5\times11+2}{11}}\)
\(=\frac{6\div\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}+\frac{57}{11}}=\frac{\left(6\div\frac{3}{5}\right)-\left(\frac{7}{6}\times\frac{6}{7}\right)}{\left(\frac{21}{5}\times\frac{10}{11}\right)+\frac{57}{11}}\)
\(=\frac{10-1}{\frac{42}{11}+\frac{57}{11}}=\frac{9}{\frac{99}{11}}=\frac{9}{9}=1\)
\(\frac{-1}{4}-\frac{3}{4}:x=\frac{-11}{36}\)
\(\frac{3}{4}:x=\frac{-1}{4}-\frac{-11}{36}\)
\(\frac{3}{4}:x=\frac{1}{18}\)
\(x=\frac{3}{4}:\frac{1}{18}\)
\(x=13,5\)
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
-1/4 - 3/4 : x = -11/36
<=> 3/4 : x = -1/4 - -11/36
<=> 3/4 : x = -1/4 + 11/36
<=> 3/4 : x = 1/18
<=> x = 3/4 : 1/18
<=> x = 27/2