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26 tháng 8 2021
1) \(2x+5\sqrt{x}-7=2\left[\left(x+\dfrac{5}{2}\sqrt{x}+\dfrac{25}{16}\right)-\dfrac{25}{16}-\dfrac{7}{2}\right]\)
\(=2\left[\left(\sqrt{x}+\dfrac{5}{4}\right)^2-\dfrac{81}{16}\right]=2\left(\sqrt{x}+\dfrac{5}{4}-\dfrac{9}{4}\right)\left(\sqrt{x}+\dfrac{5}{4}+\dfrac{9}{4}\right)=2\left(\sqrt{x}-1\right)\left(\sqrt{x}+\dfrac{7}{2}\right)\)
2) \(3x-7\sqrt{x}+4=3\left[\left(x-\dfrac{7}{3}\sqrt{x}+\dfrac{49}{36}\right)-\dfrac{49}{36}+\dfrac{4}{3}\right]\)
\(=3\left[\left(\sqrt{x}-\dfrac{7}{6}\right)^2-\dfrac{1}{36}\right]=3\left(\sqrt{x}-\dfrac{7}{6}-\dfrac{1}{6}\right)\left(\sqrt{x}-\dfrac{7}{6}+\dfrac{1}{6}\right)=3\left(\sqrt{x}-\dfrac{4}{3}\right)\left(\sqrt{x}-1\right)\)
3) \(4x-4\sqrt{x}-8=4\left[\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}-2\right]\)
\(=4\left[\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]=4\left(\sqrt{x}-\dfrac{1}{2}-\dfrac{3}{2}\right)\left(\sqrt{x}-\dfrac{1}{2}+\dfrac{3}{2}\right)=4\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)
NT
26 tháng 8 2021
| \(2x+5\sqrt{x}-7=\left(\sqrt{x}-1\right)\left(2\sqrt{x}+7\right)\) |
| \(3x-7\sqrt{x}+4=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\) |
| \(4x-4\sqrt{x}-8=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\) |
26 tháng 8 2021
`11+4sqrt6=8+2.2sqrt2.sqrt3+3=(2sqrt2+sqrt3)^2`
`11-4sqrt6=8-2.2sqrt2.sqrt3+3=(2sqrt2-sqrt3)^2`
`13+4sqrt10=8+2.2sqrt2.sqrt5+5=(2sqrt2+sqrt5)^2`
`13-4sqrt10=8-2.2sqrt2.sqrt5+5=(2sqrt2-sqrt5)^2`
26 tháng 8 2021
\(5+2\sqrt{6}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(6+2\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}\)
26 tháng 8 2021
\(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
26 tháng 8 2021
`12+2sqrt35=7+2sqrt{7.5}+5=(sqrt7+sqrt5)^2`
`9+4sqrt2=8+2.2sqrt2+1=(2sqrt2+1)^2`
`9-4sqrt2=8-2.2sqrt2+1=(2sqrt2-1)^2`
26 tháng 8 2021
\(12+2\sqrt{35}=\left(\sqrt{7}+\sqrt{5}\right)^2\)
\(9+4\sqrt{2}=\left(2\sqrt{2}+1\right)^2\)
\(9-4\left(\sqrt{2}\right)=\left(2\sqrt{2}-1\right)^2\)
NN
1
18 tháng 9 2023
`# \text {04th5}`
`a)`
\(7 \dfrac{3}{5} \div x = 5 \dfrac{4}{15} - 1 \dfrac{1}{6}\)
\(\Rightarrow 7 \dfrac{3}{5} \div x = \dfrac{41}{10}\)
\(\Rightarrow x = 7\dfrac{3}{5} \div \dfrac{41}{10}\)
\(\Rightarrow x = \dfrac{76}{41}\)
Vậy, $x = \dfrac{76}{41}$
`b)`
$x \times 2 \dfrac{2}{3} = 3 \dfrac{4}{8} + 6 \dfrac{5}{12}$
$\Rightarrow x \times \dfrac{2}{3} = \dfrac{119}{12}$
$\Rightarrow x = \dfrac{119}{12} \div \dfrac{2}{3}$
$\Rightarrow x = \dfrac{119}{8}$
Vậy, $x = \dfrac{119}{8}.$
MN
14 tháng 12 2022
1. was going
2. was attending
3. was standing
4. was riding
5. were living
6. was walking
7. was going
8. was studying / was having
9. were sitting
10. was calling / wasn't / was studying
CL
2
\(F=\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\left(đk:x\ne4\right)=\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}=\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{x+\sqrt{x}-6}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\)
\(F=\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\) \(=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{1}{3-\sqrt{x}}\)