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\(5+2\sqrt{6}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(6+2\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
`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`
\(12+2\sqrt{35}=\left(\sqrt{7}+\sqrt{5}\right)^2\)
\(9+4\sqrt{2}=\left(2\sqrt{2}+1\right)^2\)
\(9-4\left(\sqrt{2}\right)=\left(2\sqrt{2}-1\right)^2\)
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) \(4x-4\sqrt{x}-8=4\left[\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}-2\right]\)
\(=4\left[\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]=4\left(\sqrt{x}-\dfrac{1}{2}-\dfrac{3}{2}\right)\left(\sqrt{x}-\dfrac{1}{2}+\dfrac{3}{2}\right)=4\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\)
\(2x+5\sqrt{x}-7=\left(\sqrt{x}-1\right)\left(2\sqrt{x}+7\right)\) |
\(3x-7\sqrt{x}+4=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\) |
\(4x-4\sqrt{x}-8=\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\) |
`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`
\(105,48\times1,6\times\left(1,25\times5-1,25\div0,2\right)\)
\(=168,768\times\left(6,25-6,25\right)\)
\(=168,768\times0\\ =0\)
`# \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, $x = \dfrac{119}{8}.$
Ta có :
\(A=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=30+2^4\times30+2^8\times30+..2^{56}\times30\)
Vậy A chia hết cho 30 nên A cũng chia hết cho 15
hay nói cách khác A là Bội của 15
Số học sinh lớp 6A là \(120\cdot\dfrac{1}{3}=40\left(bạn\right)\)
Tổng số học sinh hai lớp 6B và 6C là 120-40=80(bạn)
Số học sinh lớp 6B là \(\dfrac{80-6}{2}=37\left(bạn\right)\)
Số học sinh lớp 6C là 37+6=43(bạn)