33/5 x 4/8 + 7/9 - 6/1
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a: \(\dfrac{-8}{3}\cdot\dfrac{15}{24}=\dfrac{-8}{24}\cdot\dfrac{15}{3}=\dfrac{-1}{3}\cdot5=-\dfrac{5}{3}\)
b: \(=\dfrac{3}{4}\cdot\dfrac{1}{-9}=\dfrac{-1}{12}\)
c: \(x=-\dfrac{7}{6}+\dfrac{5}{8}=-\dfrac{13}{24}\)
d: \(x=-\dfrac{14}{25}-\dfrac{3}{4}=\dfrac{-56-75}{100}=\dfrac{-131}{100}\)
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a: =>x=-7/6+5/8=-13/24
b: =>x=-14/25-3/4=-131/100
c: \(x=\dfrac{-33}{26}:\dfrac{-9}{13}=\dfrac{33}{26}\cdot\dfrac{13}{9}=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
d: \(x=\dfrac{4}{9}:\dfrac{5}{3}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{12}{45}=\dfrac{4}{15}\)
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a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)
\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)
=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)
b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)
Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)
c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)
\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)
=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)
d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)
\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6
=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)
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1, \(x=-\dfrac{7}{3}-\dfrac{1}{3}=-\dfrac{8}{3}\)
2, \(x=\dfrac{1}{8}-\dfrac{3}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)
3, \(x=\dfrac{6}{5}+\dfrac{4}{5}=\dfrac{10}{5}=2\)
4, \(x=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
5, \(x+\dfrac{7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}-\dfrac{7}{3}=-\dfrac{2}{3}\)
\(=x+\dfrac{1}{3}=\dfrac{-7}{3}\Leftrightarrow x=\dfrac{-8}{3}\)
\(=\dfrac{1}{8}-x=\dfrac{3}{8}\Leftrightarrow x=\dfrac{-1}{4}\)
\(=x-\dfrac{4}{5}=\dfrac{6}{5}\Leftrightarrow x=2\)
\(=\dfrac{2}{3}+x=\dfrac{7}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(=x-\dfrac{-7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{-2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{9}{6}-\dfrac{5}{7}=\dfrac{63}{42}-\dfrac{30}{42}=\dfrac{33}{42}=\dfrac{11}{14}\)
\(\dfrac{33}{12}-\dfrac{3}{8}=\dfrac{66}{24}-\dfrac{9}{24}=\dfrac{57}{24}=\dfrac{19}{8}\)
\(1-\dfrac{3}{7}=\dfrac{7}{7}-\dfrac{3}{7}=\dfrac{4}{7}\)
\(2-\dfrac{3}{4}=\dfrac{8}{4}-\dfrac{3}{4}=\dfrac{5}{4}\)
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9/5 + 4/7 + 6/5 + 3/7
= ( 9/5 + 6/5 ) + ( 4/7 + 3/7)
= 3 + 1
= 4
9/5 + 4/7 + 6/5 + 3/7 = (9/5 + 6/5) + (4/7 + 3/7)
= 15/5 + 7/7
= 3 + 1
= 4
1/2 x 45/33 x 1/9 x 11/6 = (1/2 x 1/9) x (45/33 x 11/6)
= 1/18 x 45/18
= 5/2
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1.3.77−1+3.7.99−3+7.9.1313−7+9.13.1515−9+\frac{19-13}{13.15.19}+13.15.1919−13
=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31−3.71+3.71−7.91+7.91−9.131+9.131−13.151+13.151−15.191
=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31−15.191=28595−2851=28594
b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61.(1.3.76+3.7.96+7.9.136+9.13.156+13.15.196)
làm giống như trên
c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81.(1.2.31+2.3.41+3.4.51+...+48.49.501)
=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161.(1.2.32+2.3.42+3.4.52+...+48.49.502)
=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161.(1.2.33−1+2.3.44−2+3.4.55−3+...+48.49.5050−48)
=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161.(1.21−2.31+2.31−3.41+3.41−4.51+...+48.491−49.501)
=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161.(21−24501)=161.(24501225−24501)=4900153
d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75.(1.5.87+5.8.127+8.12.157+...+33.36.407)
=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75.(1.5.88−1+5.8.1212−5+8.12.1515−8+...+33.36.4040−33)
=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75.(1.51−5.81+5.81−8.121+8.121−12.151+...+33.361−36.401)
=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75.(51−14401)=75.(1440288−14401)=28841
P/S: . là nhân nha
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6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)