(2.x-1)^2 = 10^2 -19
Mấy bạn hay mấy anh chị giải hộ em nha
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16 tháng 2 2023
\(\dfrac{x}{a}=\dfrac{m-\dfrac{x}{2}}{m}\)
\(\Rightarrow xm=a\left(m-\dfrac{x}{2}\right)\)
\(\Rightarrow xm=am-\dfrac{ax}{2}\)
\(\Rightarrow2xm=2am-ax\)
\(\Rightarrow2xm+ax=2am\)
\(\Rightarrow x\left(2m+a\right)=2am\)
\(\Rightarrow x=\dfrac{2am}{a+2m}\)
VH
1
1 tháng 8 2016
(-25).68+34.(250)
=25.68+34.250
=25.68+34.25.10
=25.(68+34).10
=25.100.10=2500
HN
0
21 tháng 3 2019
\(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024=\frac{1}{2}\left(x+y+z\right)\)
\(\Leftrightarrow2\left(\sqrt{x-2016}+\sqrt{y-2017}+\sqrt{z-2018}+3024\right)=x+y+z\)
\(\Leftrightarrow2\sqrt{x-2016}+2\sqrt{y-2017}+2\sqrt{z-2018}+6048=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x-2016}+y-2\sqrt{y-2017}+z-2\sqrt{z-2018}+6048=0\)
\(\Leftrightarrow x-2016-2\sqrt{x-2016}+1+y-2017+2\sqrt{y-2017}+1+z-2018-2\sqrt{z-2018}+1=0\)
\(\Leftrightarrow\left(\sqrt{x-2016}-1\right)^2+\left(\sqrt{y-2017}-1\right)^2+\left(\sqrt{z-2018}-1\right)^2=0\)
\(ĐK:x\ge2016;y\ge2017;z\ge2018\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}-1=0\\\sqrt{y-2017}-1=0\\\sqrt{z-2018}-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{x-2016}=1\\\sqrt{y-2017}=1\\\sqrt{z-2018}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2017\\y=2018\\z=2019\end{cases}}}\)
NT
0
14 tháng 7 2019
a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=12\)
\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12\)
\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0\)
\(\Leftrightarrow-12x-27=0\)
\(\Leftrightarrow x=\frac{-9}{4}\)
b) xem lại đề
c) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x-3\right)^2=1\)
\(\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0\)
\(\Leftrightarrow x^3-28-x^3+6x^2-9x=0\)
\(\Leftrightarrow6x^2-9x-28=0\)
\(\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}\)
d) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(\Leftrightarrow12x+15=0\)
\(\Leftrightarrow x=\frac{-5}{4}\)
14 tháng 7 2019
Theo giả thiết:
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Dễ thấy \(VT\ge0\forall a;b;c\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)(đpcm)
1 tháng 4 2017
Vì nó bé hơn thì nó bé hơn
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mấy chế lm thiếu hết òi , số chẵn sẽ chia ra lm 2 trường hợp
\(\left(2.x-1\right)^2=10^2-19\)
\(\left(2x-1\right)^2=100-19\)
\(\left(2x-1\right)^2=81\)
\(\left(2x-1\right)^2=9^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
\(\left(2x-1\right)^2=10^2-19\)
\(\left(2x-1\right)^2=100-19\)
\(\left(2x-1\right)^2=81\)
\(\left(2x-1\right)^2=9^2\)
\(2x-1=9\)
\(2x=9+1\)
\(2x=10\)
\(x=10:2\)
\(\Rightarrow x=5\)