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a) \(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(\Leftrightarrow A=-5,13:\left(5\dfrac{5}{28}-\dfrac{85}{36}+1\dfrac{16}{63}\right)\)
\(\Leftrightarrow A=-5,13:\left(\dfrac{355}{126}+1\dfrac{16}{63}\right)\)
\(\Leftrightarrow A=-5,13:\dfrac{57}{14}\)
\(\Leftrightarrow A=\dfrac{-63}{50}\)
b) \(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)
\(\Leftrightarrow B=\left(\dfrac{19}{3}+19,5.\dfrac{3}{13}\right).\dfrac{2}{3}\)
\(\Leftrightarrow B=\left(\dfrac{19}{3}+\dfrac{9}{2}\right).\dfrac{2}{3}\)
\(\Leftrightarrow B=\dfrac{65}{6}.\dfrac{2}{3}\)
\(\Leftrightarrow B=\dfrac{65}{9}\)
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
\(\left|\dfrac{x}{2015}+\dfrac{x}{2016}\right|=\left|\dfrac{x}{2016}+\dfrac{x}{2017}\right|\)
\(\Rightarrow\left|x\right|.\left|\dfrac{1}{2015}+\dfrac{1}{2016}\right|=\left|x\right|.\left|\dfrac{1}{2016}+\dfrac{1}{2017}\right|\)
\(\Rightarrow\left|x\right|.\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)=\left|x\right|.\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
Từ đó \(\Rightarrow\)
\(\left|x\right|.\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left|x\right|.\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)=0\)
\(\Rightarrow\left|x\right|.\left[\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\right]=0\)
\(\Rightarrow\left|x\right|=0:\left[\left(\dfrac{1}{2015}+\dfrac{1}{2016}\right)-\left(\dfrac{1}{2016}+\dfrac{1}{2017}\right)\right]\)
\(\Rightarrow\left|x\right|=0\Rightarrow x=0\)
Vậy \(x=0\)
\(\dfrac{\left|x-7\right|}{\left|x-4\right|}=\dfrac{\left|x-1\right|}{\left|x-4\right|}\)
Ta xét: \(\left|x-7\right|\ge0;\left|x-4\right|\ge0\)
\(\Rightarrow\dfrac{\left|x-7\right|}{\left|x-4\right|}\ge0\)
Ta lại xét: \(\left|x-1\right|\ge0;\left|x-4\right|\ge0\)
\(\Rightarrow\dfrac{\left|x-1\right|}{\left|x-4\right|}\ge0\)
\(\dfrac{\left|x-7\right|}{\left|x-4\right|}=\dfrac{\left|x-1\right|}{\left|x-4\right|}\)
\(\Rightarrow\dfrac{x-7}{x-4}=\dfrac{x-1}{x-4}\)
\(\Rightarrow x-7=x-1\)
=> \(x\in\varnothing\)
Ta có:\(\frac{x+5}{x+3}< 1\)
\(\Rightarrow\frac{x+3}{x+3}+\frac{2}{x+3}< 1\)
\(\Rightarrow1+\frac{2}{x+3}< 1\)
\(\Rightarrow\frac{2}{x+3}< 0\)
\(\Rightarrow x+3< 0\)
\(\Rightarrow x< -3\)
b,c tương tự nha@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{1}{3}=\dfrac{x+y}{\left(x+y\right)+2\left(z+t\right)}\)
\(\Rightarrow\left(x+y\right)+2\left(z+t\right)=3\left(x+y\right)\)
\(\Rightarrow2\left(z+t\right)=2\left(x+y\right)\Rightarrow\dfrac{x+y}{z+t}=1\)
Chứng minh tương tự ta được:
\(\dfrac{y+z}{x+t}=1;\dfrac{z+t}{x+y}=1;\dfrac{t+x}{y+z}=1\)
\(\Rightarrow P=1+1+1+1=4\)
+Xét x+y+z+t=0
\(\Rightarrow\)\(\left\{{}\begin{matrix}z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\\x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\end{matrix}\right.\)
Khi đó M=-4
+Xét x+y+z+t\(\ne\)0
ADTC dãy tỉ số bằng nhau ta có
\(\dfrac{x}{y+z+t}\)=\(\dfrac{y}{x+y+t}\)=\(\dfrac{z}{x+y+t}\)=\(\dfrac{z}{x+y+t}\)=\(\dfrac{x+y+z+t}{3.\left(x+y+z+t\right)}\)=\(\dfrac{1}{3}\)
+Với\(\dfrac{x}{y+z+t}\)=\(\dfrac{1}{3}\)
\(\Rightarrow\)3x=y+z+t
\(\Rightarrow\)4x=x+y+z+t
Chứng minh tương tự ta có
4y=x+y+z+t
4z=x+y+z+t
4t=x+y+z+t
Do đó x=y=z=t
Khi đó M=4
\(\Leftrightarrow x^2=4\cdot25=100\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
Vậy có 2 gt x thỏa