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- SABC=\(\dfrac{1}{2}\)AB.AC=\(\dfrac{1}{2}\).4.6=12(cm2)
- Xét tam giác ABC vuông tại A có:
BC2=AB2+AC2 (định lí Py-ta-go)
=>BC2=42+62=52
=>BC=\(\sqrt{52}\)(cm)
- Xét tam giác ABC có:
AD là đường phân giác của góc A (gt)
=>\(\dfrac{AB}{AC}=\dfrac{BD}{DC}\)(t/c đường phân giác)
=>\(\dfrac{AB+AC}{AC}=\dfrac{BC}{DC}\)
=>\(\dfrac{4+6}{6}=\dfrac{\sqrt{52}}{DC}\)
=>DC=\(\dfrac{6\sqrt{13}}{5}\)
- Ta có: DE vuông góc với AB (gt) ; AC vuông góc với AB (tam giác ABC vuông tại A).
=>DE//AC.
- Xét tam giác ABC có:
DE//AC (cmt)
=>\(\dfrac{AE}{AB}=\dfrac{CD}{BC}\)(định lí Ta-let)
=>\(\dfrac{AE}{4}=\dfrac{\text{}\text{}\dfrac{6\sqrt{13}}{5}}{\sqrt{52}}\)
=>AE=2,4 (cm)
- Ta có: Góc EAF=900(Tam giác ABC vuông tại A)
Góc AED =900(DE vuông góc với AB tại E)
Góc AFD=900(DF vuông góc với AC tại F)
=>DEAF là hình chữ nhật.
Mà AD là phân giác của góc EAF (gt)
=>DEAF là hình vuông.
=>AE=AF=2,4 (cm)
=> SAEF=\(\dfrac{1}{2}\)AE.AF=\(\dfrac{1}{2}\).2,4.2,4=2,88 (cm2)
- SBEFC=SABC-SAEF=12-2,88=9,12 (cm2).
-->Chọn câu A
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\(x\cdot\dfrac{3}{7}-x\cdot\dfrac{1}{2}=\dfrac{3}{5}\)
\(x\left(\dfrac{3}{7}-\dfrac{1}{2}\right)=\dfrac{3}{5}\)
\(x\cdot\dfrac{-1}{14}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{14}\)
\(x=\dfrac{-42}{5}\)
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14:A (có did)
15:B (tug of war có nghĩa là kéo co)
16:B ( vì A mang nghĩa đi chơi cùng gd nhưng đề bải hỏi là quê hương nên chọn B)
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Students should play sports because it helps strengthen bones and muscles. Firstly, physical activity is very good for children's bone development, helps keep bone density at a high level, and reduces the risk of osteoporosis. Secondly, When we exercise, the brain releases a chemical that creates feelings of happiness and helps relieve stress. Beside, participating in a team sport will give your child the opportunity to meet many new friends, with different ages and personalities. As you see, sport not only brings health and dynamism, sports also help children develop teamwork, cooperation skills, discipline, ... and focus better during class!
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\(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{\left(x+y\right)^2-z^2-2xy}+\dfrac{xz}{\left(x+z\right)^2-y^2-2xz}+\dfrac{yz}{\left(y+z\right)^2-x^2-2yz}\)
\(=\dfrac{xy}{\left(x+y-z\right)\left(x+y+z\right)-2xy}+\dfrac{xz}{\left(x+z-y\right)\left(x+z+y\right)-2xz}+\dfrac{yz}{\left(y+z-x\right)\left(y+x+z\right)-2yz}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)
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\(\dfrac{x-2}{4}=\dfrac{5+x}{3}\)
⇒3.(x+2)=4.(5+x)
3x+6=20+x
3x+6-x=20
2x+6=20
2x=14
x=7