34+2.(x-5)mũ 2=66
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\(2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}\)
Đặt \(\dfrac{x}{5}=\dfrac{y}{2}=k\Rightarrow x=5k;y=2k\)
\(x^2+2y^2=66\\ \Rightarrow25k^2+8k^2=66\\ \Rightarrow33k^2=66\\ \Rightarrow k^2=2\Rightarrow\left[{}\begin{matrix}k=\sqrt{2}\\k=-\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\sqrt{2};y=2\sqrt{2}\\x=-5\sqrt{2};y=-2\sqrt{2}\end{matrix}\right.\)
Đặt \(\dfrac{x}{5}=\dfrac{y}{2}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)
Ta có: \(x^2+2y^2=66\)
\(\Leftrightarrow33k^2=66\)
\(\Leftrightarrow k^2=2\)
Trường hợp 1: \(k=\sqrt{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\sqrt{2}\\y=2\sqrt{2}\end{matrix}\right.\)
Trường hợp 2: \(k=-\sqrt{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\sqrt{2}\\y=-2\sqrt{2}\end{matrix}\right.\)
23.27.24=23+7+4=214
23.24:25=23+4-5=27
(-3)+(-125)+(-25)
=-128+(-25)
=-153
25+(-38)=25-38=-13
126+159=285
120+(-135)+200
=120-135+200
=-15+200
=185
2^3x2^7=2^10x2^4=2^14=16384
2^3x2^4=2^7:2^5=2^2=4
1300-(150.2+(900+90):9)=1300-(150.2+990:9)=1300-(300+11)=1300-311=989
(-3) + (-125)+(-25)= -(3+125+25)= -153
25 + (-38)= -(38-25)= - 13
(126)+159=285
(120)+(-135)+200= -(135-120) + 200= (-15) + 200= +(200-15)=185
Đặt \(\dfrac{x}{4}=\dfrac{y}{7}=k\Rightarrow x=4k.y=7k\)
Ta có:\(y^2-x^2=66\\ \Rightarrow\left(7k\right)^2-\left(4k\right)^2=66\\ \Rightarrow49k^2-16k^2=66\\ \Rightarrow33k^2=66\\ \Rightarrow k^2=2\\ \Rightarrow\left[{}\begin{matrix}k=\sqrt{2}\\k=-\sqrt{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\sqrt{2};y=7\sqrt{2}\\x=-4\sqrt{2};y=-7\sqrt{2}\end{matrix}\right.\)
a) 27 + 34 + 66 = 27 + (34 + 66) = 27 + 100 = 127 | b) 7 x 5 x 2 = 7 x (5 x 2) = 7 x 10 = 70 |
`34+2(x-5)^2=66`
`=>2(x-5)^2=66-34`
`=>2(x-5)^2=32`
`=>(x-5)^2=32:2`
`=>(x-5)^2=16=(4)^2=(-4)^2`
`=>` $\left[ \begin{array}{l}x-5=4\\x-5=-4\end{array} \right.$
`=>` $\left[ \begin{array}{l}x=9\\x=1\end{array} \right.$
Vậy x=9 hoặc x=1
Ta có: \(34+2\left(x-5\right)^2=66\)
\(\Leftrightarrow2\left(x-5\right)^2=32\)
\(\Leftrightarrow\left(x-5\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;9\right\}\)