Tính:
a) \(\sqrt{a^2}\) với a= 6,5 ; -0,1 b)\(\sqrt{a^4}\) với a= 3; -0,1 c)\(\sqrt{a^6}\) với a= -2; 0,1
Tính:
a) \(\sqrt{a^2}\) với a= 6,5 ; -0,1 b)\(\sqrt{a^4}\) với a= 3; -0,1 c)\(\sqrt{a^6}\) với a= -2; 0,1
a) \(\sqrt{a^2}=\left|a\right|\)
Khi \(a=6,5\Rightarrow\sqrt{a^2}=\left|6,5\right|=6,5\)
Khi \(a=-0,1\Rightarrow\sqrt{a^2}=\left|-0,1\right|=0,1\)
b) \(\sqrt{a^4}=\left|a^2\right|=a^2\)
Khi \(a=3\Rightarrow\sqrt{a^4}=3^2=9\)
Khi \(a=-0,1\Rightarrow\sqrt{a^4}=\left(-0,1\right)^2=0,01\)
c) \(\sqrt{a^6}=\left|a^3\right|\)
Khi \(a=-2\Rightarrow\sqrt{a^6}=\left|\left(-2\right)^3\right|=\left|-8\right|=8\)
Khi \(a=0,1\Rightarrow\sqrt{a^6}=\left|0,1^3\right|=0,001\)
Giúp ;-; Giải chị tiết cho mình hiểu đc ko ạ ;-;
`1)\sqrt{[3a]/2}.\sqrt{[8a^3]/27}` `(a >= 0)`
`=\sqrt{[3a.8a^3]/[2.27]}`
`=\sqrt{[24a^4]/54}=\sqrt{[(2 a^2)^2 .6]/[3^2 .6]}=[2a^2]/3`
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`2)\sqrt{2x^3}.\sqrt{8xy}.\sqrt{9y^5}` `(x >= 0,y >= 0)`
`=\sqrt{2x^3 .8xy.9y^5}`
`=\sqrt{144x^4y^6}`
`=\sqrt{(12x^2y^3)^2}=|12x^2y^3|=12x^2y^3` (Vì `x >= 0,y >= 0`)
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`3)[x+\sqrt{xy}]/[y+\sqrt{xy}]` `(x > 0,y > 0)`
`=[\sqrt{x}(\sqrt{x}+\sqrt{y})]/[\sqrt{y}(\sqrt{x}+\sqrt{y})]`
`=\sqrt{x}/\sqrt{y}`
Rút gọn biểu thức sau:
a)\(\sqrt{50^2}-14^2\) b)\(\sqrt{34^2-16^2}\) c)\(\sqrt{1,5}.\sqrt{\dfrac{2}{3}}\) d)\(\sqrt{1\dfrac{1}{8}}.\sqrt{0,72}\)
(giúp mình với giải chi tiết xíu ạ)
a: =50-196=-146
b: \(=\sqrt{\left(34-16\right)\left(34+16\right)}=\sqrt{18\cdot50}=30\)
c: \(=\sqrt{\dfrac{3}{2}\cdot\dfrac{2}{3}}=1\)
d: \(=\sqrt{\dfrac{9}{8}\cdot\dfrac{18}{25}}=\sqrt{\dfrac{9}{25}\cdot\dfrac{9}{4}}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
\(\sqrt{x-2\sqrt{x-1}}=3\)
\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}=3\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=3\)
\(\Leftrightarrow\sqrt{x-1}-1=3\)
\(\Leftrightarrow\sqrt{x-1}=4\)
=>x-1=16
hay x=17
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\) tính ghi rõ bc giải
\(=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
A và b ko âm giải thích a lớn hơn hoặc bằng 0 và b lớn hơn hoặc bằng 0
Giá trị không âm tức là giá trị =0 hoặc lớn hơn 0, ký hiệu là $\geq 0$
Căn (3/2 - 1/2 căn 5)
\(\sqrt{\dfrac{3-\sqrt{5}}{2}}=\sqrt{\dfrac{6-2\sqrt{5}}{4}}=\dfrac{\sqrt{5}-1}{2}\)
A=13−2√42
`A=13-2\sqrt{42}`
`A=(\sqrt{7})^2-2\sqrt{7}.\sqrt{6}+(\sqrt{6})^2`
`A=(\sqrt{7}-\sqrt{6})^2`
A=13−2√42
\(=\left(\sqrt{7}\right)^2-2\sqrt{7}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{7}-\sqrt{6}\right)^2\)
giup vs mn ơi mik đag gấp
Do pt có 2 nghiệm \(x_1,x_2\) nên ta có :
\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{2}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)
Ta có :
\(P=x_1\left(3+x_2\right)+x_2\left(3+x_1\right)+3x^2_1+3x^2_2-10\)
\(=3x_1+x_1x_2+3x_2+x_1x_2+3\left(x_1^2+x_2^2\right)-10\)
\(=3\left(x_1+x_2\right)+2x_1x_2+3\left(x^2_1+x^2_2\right)-10\)
\(=3S+2P+3\left(S^2-2P\right)-10\)
\(=3.\left(-\dfrac{5}{2}\right)+2.\left(-\dfrac{1}{2}\right)+3\left(\left(-\dfrac{5}{2}\right)^2-2\left(-\dfrac{1}{2}\right)\right)-10\)
\(=\dfrac{13}{4}\)
Vậy \(P=\dfrac{13}{4}\)
Giúp em vs 1,2 giải denta nha🙁
a.\(5x^2-4x-1=0\)
\(\Delta=\left(-4\right)^2-4.5.\left(-1\right)=16+20=36>0\)
=> pt có 2 nghiệm phân biệt
\(\left\{{}\begin{matrix}x_1=\dfrac{4+\sqrt{36}}{10}=1\\x_2=\dfrac{4-\sqrt{36}}{10}=-\dfrac{1}{5}\end{matrix}\right.\)
b.\(7x^2-2\sqrt{7}x+1=0\)
\(\Delta=\left(-2\sqrt{7}\right)^2-4.1.7\)
\(=28-28=0\)
=> pt có nghiệm kép
\(x_1=x_2=\dfrac{2\sqrt{7}}{2.7}=\dfrac{\sqrt{7}}{7}\)
a, \(\Delta=\left(-4\right)^2-4.5.\left(-1\right)=16+20=36>0\)
\(x_1=\dfrac{4+6}{10}=1\\ x_2=\dfrac{4-6}{10}=\dfrac{-1}{5}\)
\(b,\Delta=\left(-2\sqrt{7}\right)^2-4.7.1=28-28=0\)
\(\Rightarrow\) pt có nghiệm kép là: \(\dfrac{2\sqrt{7}}{14}=\dfrac{\sqrt{7}}{7}\)
\(c,4x^2-2=0\\ \Leftrightarrow2x^2-1=0\\ \Leftrightarrow x^2=\dfrac{1}{2}\\ \Leftrightarrow x=\pm\dfrac{\sqrt{2}}{2}\)
\(d,\left\{{}\begin{matrix}4x+y=3\\2x-3y=8\end{matrix}\right.\\ \Leftrightarrow \left\{{}\begin{matrix}y=3-4x\\2x-3\left(3-4x\right)=8\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}y=3-4x\\2x-9+12x=8\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}y=3-4x\\14x=17\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}y=3-4.\dfrac{17}{14}\\x=\dfrac{17}{14}\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}y=-\dfrac{13}{7}\\x=\dfrac{17}{14}\end{matrix}\right.\)