cho e hỏi bài này với ạ
\(\dfrac{2}{36}+\dfrac{1}{6}+\dfrac{5}{12}\)
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\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
\(A=1-\dfrac{1}{2}+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}\)
\(=4-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}\right)\)
\(=4-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)\)
=4-4/5=16/5
Bài 2:
1: \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)
=>\(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}=\dfrac{1}{12}+\dfrac{10}{12}=\dfrac{11}{12}\)
=>x=11
2: \(\dfrac{2}{3}-1\dfrac{4}{15}x=-\dfrac{3}{5}\)
=>\(\dfrac{2}{3}-\dfrac{19}{15}x=-\dfrac{3}{5}\)
=>\(\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10+9}{15}=\dfrac{19}{15}\)
=>\(x=\dfrac{19}{15}:\dfrac{19}{15}=1\)
3: \(\dfrac{\left(-3\right)^x}{81}=-27\)
=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)
=>x=7
4: \(\left|x+0,237\right|=0\)
=>x+0,237=0
=>x=-0,237
5: \(\left(x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
6: \(\left|2x-1\right|=5\)
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
7: \(\left(x-1\right)^3=-\dfrac{8}{27}\)
=>\(\left(x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
=>\(x-1=-\dfrac{2}{3}\)
=>\(x=-\dfrac{2}{3}+1=\dfrac{1}{3}\)
8: \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
=>\(\dfrac{5}{3}:\dfrac{x}{4}=20\)
=>\(\dfrac{20}{3x}=20\)
=>3x=20/20=1
=>\(x=\dfrac{1}{3}\)
9: \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:2\dfrac{2}{3}\)
=>\(\dfrac{\dfrac{8}{3}}{x}=\dfrac{\dfrac{16}{9}}{\dfrac{8}{3}}\)
=>\(\dfrac{16}{9}\cdot x=\dfrac{8}{3}\cdot\dfrac{8}{3}=\dfrac{64}{9}\)
=>16x=64
=>x=64/16=4
Bài 3:
1: Ta có: x-24=y
=>x-y=24
mà \(\dfrac{x}{7}=\dfrac{y}{3}\)
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)
=>\(x=6\cdot7=42;y=6\cdot3=18\)
2: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\)
mà x-y=48
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{x-y}{5-7}=\dfrac{48}{-2}=-24\)
=>\(x=-24\cdot5=-120;y=-24\cdot7=-168;z=-24\cdot2=-48\)
3: \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\)
mà x-y=4009
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}=\dfrac{x-1+3-y}{2005+2006}=\dfrac{4009+2}{4011}=1\)
=>\(x-1=2005;3-y=2006\)
=>x=2005+1=2006; y=3-2006=-2003
5: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)
mà 2x+3y-z=-14
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{2\cdot3+3\cdot5-7}=\dfrac{-14}{14}=-1\)
=>\(x=-3;y=-5;z=-7\)
Bạn tách ra từng CH khác nhau đi nhé. Gộp 1 trong tất cả rất khó nhìn và lâu.
ĐKXĐ: \(x\notin\left\{0;-9\right\}\)
Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)
Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)
\(\Leftrightarrow9x^2+81x+180=0\)
\(\Leftrightarrow x^2+9x+20=0\)
\(\Leftrightarrow x^2+4x+5x+20=0\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-4;-5}
\(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}\Rightarrow x=\dfrac{1}{10}\)
a)
\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-12\times\dfrac{144}{1}\)
\(=-1728\)
b)
\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)
\(=3-17\)
\(=-14\)
a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)
\(\dfrac{2}{36}+\dfrac{1}{6}+\dfrac{5}{12}=\dfrac{2}{36}+\dfrac{6}{36}+\dfrac{15}{36}=\dfrac{2+6+15}{36}=\dfrac{23}{36}\)