2018*x(x-2018/7)=0
Những câu hỏi liên quan
a, (x+2018)(1/2+2/7)=(x+2018).(1/5+1/6)
b, 7(x-1)+2x(x-1)=0
a) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)=\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\)
\(\Leftrightarrow\) \(\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}\right)-\left(x+2018\right)\left(\frac{1}{5}+\frac{1}{6}\right)\) = 0
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2}+\frac{2}{7}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+2018=0\)
\(\Leftrightarrow x=-2018\)
b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7+2x\right)=0\)
\(\Leftrightarrow\) x - 1 = 0 hoặc 7 + 2x = 0
1) x - 1 = 0 \(\Leftrightarrow\) x = 1
2) 7 + 2x = 0 \(\Leftrightarrow\) -3,5
Vậy: x = 1; -3,5
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b) \(7\left(x-1\right)+2x\left(x-1\right)=0\)
=> \(\left(x-1\right).\left(7+2x\right)=0\)
=> \(\left\{{}\begin{matrix}x-1=0\\7+2x=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0+1\\2x=0-7=-7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1\\x=\left(-7\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\\x=-\frac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;-\frac{7}{2}\right\}.\)
Chúc bạn học tôt!
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Tìm x, y
| x - 2017 | + | y - 2018 | ≤ 0
3| x - y |5 + 10| y + 2/3 |7 ≤ 0
1/2(3/4x - 1/2)2018 + 2017/2018|4/5 y+ 6/25| ≤ 0
2017 |2x - y | 2018 + 2018 | y - 4 |2017 ≤ 0
Tìm x,y thoả mãn
/x-2017/+/y-2018/ <=0
/3.x-y/^5+10./y+2/3/^7 <=0
c,1/2.(3/4.x-1/2)^2018+2017/2019./4/5.y+6/25/<=0
d,2017./2x-y/^2018+2018./y-4/^2017<=0
giúp em vs m.n ưi,mai em nộp ùi
a: =>x-2017=0 và y-2018=0
=>x=2017; y=2018
b: =>3x-y=0 và y+2/3=0
=>y=-2/3 và 3x=-2/3
=>x=-2/9 và y=-2/3
c: =>3/4x-1/2=0 và 4/5y+6/25=0
=>x=2/3 và y=-3/10
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Tìm giá trị nhỏ nhất
P = 2018/x^2+2x+2017
Q = a^2018+2017/a^2018+2015
A = (x-3y)^2020+(y-2018)^2018
B = (x+y-5)^8+(x-2y)^4+2016
C = \x-2017\+\x-2018\
D = \x-2010\+\x-2011\+\x+2012\
a)A=/x+7/+/x^2-169/-/x-2018/
b)B=[2018/2+2018/3+2028/4+.....+2019/2018]:[1/2018+2/2017+3/2016+......+2018]
sqrt{x^2+2018}+xsqrt{x^2}x
sqrt{x^2+2018}-x0 sqrt{x^2+2018}-xkhác 0 (left(sqrt{x^2+2018}-xright)left(sqrt{x^2+2018}+xright)left(sqrt{y^2+2018}+yright)2018left(sqrt{x^2+2018}-xright) 2018left(sqrt{y^2+2018}+yright) 2018left(sqrt{x^2+2018}-xright) sqrt{y^2+2018}+ysqrt{x^2+2018}-xChứng minh tương tự sqrt{x^2+2018}+xsqrt{y^2+2018}-yCộng 2 cái vào. Khử được hạng tử. suy ra đc x+y0 rồi tự làm cưng e nhé
Đọc tiếp
\(\sqrt{x^2+2018}+x>\sqrt{x^2}>=x \)
=> \(\sqrt{x^2+2018}-x>0\)
=> \(\sqrt{x^2+2018}-x\)khác 0
=> (\(\left(\sqrt{x^2+2018}-x\right)\left(\sqrt{x^2+2018}+x\right)\left(\sqrt{y^2+2018}+y\right)=2018\left(\sqrt{x^2+2018}-x\right)\)
<=> 2018\(\left(\sqrt{y^2+2018}+y\right)\)= 2018\(\left(\sqrt{x^2+2018}-x\right)\)
<=> \(\sqrt{y^2+2018}+y=\sqrt{x^2+2018}-x\)
Chứng minh tương tự => \(\sqrt{x^2+2018}+x=\sqrt{y^2+2018}-y\)
Cộng 2 cái vào. Khử được hạng tử. suy ra đc x+y=0 rồi tự làm cưng e nhé
tìm x biết (5^2018+5^2018+5^2018+5^2018)+5^2018-5x=0
(5^2018+5^2018+5^2018+5^2018) + 5^2018 -5x=0
5^2018+5^2018+5^2018+5^2018+5^2018-5x =0
5(5^2018)-5x =0
5x =5(5^2018)-0
5x =5(5^2018)
Suy ra x= 5^2018
Vậy: x= 5^2018
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Cho x, y, z >0, x+y+z=2018. C/m biểu thức sau không phụ thuộc vào x:
m = x.\(\sqrt{\frac{\left(y^2+2018\right).\left(z^2+2018\right)}{x^2+2018}}+y.\sqrt{\frac{\left(x^2+2018\right).\left(z^2+2018\right)}{y^2+2018}}+z.\sqrt{\frac{\left(x^2+2018\right).\left(y^2+2018\right)}{z^2+2018}}\)
5x(x-2018)-x+2018=0
x^3-2x=0
Câu a :
\(5x\left(x-2018\right)-x+2018=0\)
\(5x\left(x-2018\right)-x+2018=0\)
\(\Leftrightarrow5x\left(x-2018\right)-\left(x-2018\right)=0\)
\(\Leftrightarrow\left(x-2018\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2018=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2018\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{5}\) hoặc \(x=2018\)
Câu b :
\(x^3-2x=0\)
\(\Leftrightarrow x\left(x^2-2\right)=0\)
\(\Leftrightarrow x\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\sqrt{2}=0\\x+\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Vậy \(x=-\sqrt{2}\) ; \(x=0\) hoặc \(x=\sqrt{2}\)
Wish you study well !!
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