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\(\dfrac{5}{13}+\dfrac{5}{17}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
\(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{5}{17}\)
\(=1+\left(-1\right)+\dfrac{5}{17}\)
\(=0+\dfrac{5}{17}=\dfrac{5}{7}\)
\(0+\dfrac{5}{17}=\dfrac{5}{17}\)
Đs pải như này ms đúng chứ Hằng!?
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
3)\(\dfrac{-41}{32}\left(\dfrac{15}{8}-\dfrac{16}{41}\right)+\dfrac{15}{8}\left(\dfrac{41}{32}-\dfrac{8}{3}\right)\)
=\(\dfrac{-41}{32}.\dfrac{15}{8}-\dfrac{-41}{32}.\dfrac{16}{41}+\dfrac{15}{8}.\dfrac{41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(\left(\dfrac{-41}{32}.\dfrac{15}{8}+\dfrac{15}{8}.\dfrac{41}{32}\right)+\dfrac{-16}{41}.\dfrac{-41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(0+\dfrac{1}{2}-5=\dfrac{-9}{2}\)
4)\(\dfrac{13}{29}\left(\dfrac{29}{5}-\dfrac{45}{8}\right)-\dfrac{45}{8}\left(\dfrac{9}{8}-\dfrac{13}{29}\right)\)
=\(\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{9}{8}-\dfrac{45}{8}.\dfrac{13}{29}\)
=\(\left(\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{13}{29}\right)-\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{9}{8}\)
=\(0-\dfrac{13}{5}-\dfrac{405}{64}=\dfrac{-2857}{320}\)
a. 857+3,15+127+4,35
=\(\dfrac{61}{7}+\dfrac{63}{20}+\dfrac{9}{7}+\dfrac{87}{20}\)
=\(\left(\dfrac{61}{7}+\dfrac{9}{7}\right)+\left(\dfrac{63}{20}+\dfrac{87}{20}\right)\)
=\(10+\dfrac{15}{2}\)
=\(\dfrac{35}{2}\)
b. (4523−225+7713)−(3523−6613)
=\(4\dfrac{5}{23}-2\dfrac{2}{5}+7\dfrac{7}{13}-3\dfrac{5}{23}+6\dfrac{6}{13}\)
=\(\left(4\dfrac{5}{23}-3\dfrac{5}{23}\right)+\left(7\dfrac{7}{13}+6\dfrac{6}{13}\right)-2\dfrac{2}{5}\)
=\(1+14-\dfrac{12}{5}\)
=15-\(\dfrac{12}{5}\)
=\(\dfrac{63}{5}\)
Câu C khó khó mình chưa giải được !!!
Giải:
a) \(A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{-20}{41}+\dfrac{5}{13}+\dfrac{-21}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{5}{13}+\dfrac{-21}{41}+\dfrac{-20}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}\left(\dfrac{5}{7}+1\right)+\dfrac{-41}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{12}{7}+\left(-1\right)\)
\(\Leftrightarrow A=\dfrac{60}{91}+\left(-1\right)=-\dfrac{31}{91}\)
Vậy ...
b) \(B=\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\)
\(\Leftrightarrow B=\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{11}\)
Vậy ...
c) \(C=\dfrac{-2}{3}+\dfrac{-5}{7}+\dfrac{2}{3}+\dfrac{-2}{7}\)
\(\Leftrightarrow C=\left(\dfrac{-2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{7}+\dfrac{-5}{7}\right)\)
\(\Leftrightarrow C=0+\left(-1\right)=-1\)
Vậy ...
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Quy đồng ( đây ngu toán không logic )
a)\(\dfrac{3}{6};\dfrac{2}{6};\dfrac{4}{6}\Rightarrow\dfrac{4}{6}>\dfrac{3}{6}>\dfrac{2}{6}\)
b)\(\dfrac{16}{9};\dfrac{24}{13}=\dfrac{208}{117};\dfrac{216}{117}\Rightarrow\dfrac{216}{117}>\dfrac{208}{117}\)
c)(Trời ơi cái đề bài)
h)\(\dfrac{27}{82};\dfrac{26}{75}=\dfrac{2025}{6150};\dfrac{2132}{6150}\Rightarrow\dfrac{2025}{6150}< \dfrac{2132}{6150}\)
d)(Trời ơi giống câu c)
i)\(\dfrac{-49}{78};\dfrac{64}{-95}=\dfrac{-4655}{7410};\dfrac{4992}{7410}\Rightarrow\dfrac{-4655}{7410}< \dfrac{4992}{7410}\)
P/s : Tự kết luận mỗi câu
a,\(\dfrac{1}{2}=\dfrac{1.3}{2.3}=\dfrac{3}{6}\),\(\dfrac{1}{3}=\dfrac{1.2}{3.2}=\dfrac{2}{6}\),\(\dfrac{2}{3}=\dfrac{2.2}{3.2}=\dfrac{4}{6}\)
vì có mẫu chung là 6 nên ta so sánh tử\(\Rightarrow\)ta so sánh 3,2,4
vì 2<3<4\(\Rightarrow\)\(\dfrac{2}{6}< \dfrac{3}{6}< \dfrac{4}{6}\Rightarrow\dfrac{1}{3}< \dfrac{1}{2}< \dfrac{2}{3}\)