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Câu a) bạn tham khảo tại đây nhé: Câu hỏi của Hằng Thanh.
Chúc bạn học tốt!
\(\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{2}{9}+\frac{1}{36}\right)+\frac{1}{72}=1-1+\frac{1}{72}=\frac{1}{72}\)
\(A=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{72}\\ =\frac{1}{9}-\frac{7}{9}+\frac{2}{3}+\frac{1}{72}=-\frac{2}{3}+\frac{2}{3}+\frac{1}{72}=\frac{1}{72}\)
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
\(=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)+\left(-\frac{3}{4}-\frac{2}{9}-\frac{1}{36}\right)+\frac{1}{64}\)
= 1 + -1 + 1/64
= 0 +1/64
= 1/64
1/3*(-3)/4 - -3/5+ 1/72*(-2)/9*(-1)/36 + 1/15
= 1/72
Đảm bảo 100 % là đúng
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
a: \(2\cdot\sqrt{\frac{25}{16}}-3\cdot\sqrt{\frac{49}{36}}+4\cdot\sqrt{\frac{81}{64}}\)
\(=2\cdot\frac54-3\cdot\frac76+4\cdot\frac98\)
\(=\frac52-\frac72+\frac92=\frac{5-7+9}{2}=\frac72\)
b: \(\left(3\sqrt2\right)^2-\left(4\cdot\sqrt{\frac12}\right)^2+\frac{1}{16}\cdot\left(\sqrt{\frac34}\right)^2\)
\(=18-16\cdot\frac12+\frac{1}{16}\cdot\frac34\)
\(=18-8+\frac{3}{64}=10+\frac{3}{64}=\frac{643}{64}\)
c: \(\frac23\cdot\sqrt{\frac{81}{16}}-\frac34\cdot\sqrt{\frac{64}{9}}+\frac75\cdot\sqrt{\frac{25}{196}}\)
\(=\frac23\cdot\frac94-\frac34\cdot\frac83+\frac75\cdot\frac{5}{14}\)
\(=\frac12\cdot3-\frac84+\frac{7}{14}=\frac32+\frac12-2=0\)