giúp mình với
x-2/8=-2/2-x.1/3
mình cảm ơn trước
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1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
(x+1)(x+2)(x+3)(x+4)
= (x+1)(x+4)(x+2)(x+3)
= (x2+5x+4)(x2+5x+6)
=(x2+5x+4)2+2(x2+5x+4)+1-1
= (x2+5x+5)2-1
Vì (x2+5x+5)2 luôn lớn hơn hoặc bằng 0 với mọi x=> (x2+5x+5)2-1 luôn lớn hơn hoặc bằng -1 với mọi x
=> GTNN của (x+1)(x+2)(x+3)(x+4) là -1 khi và chỉ khi x = \(\sqrt{1,25}\)-2,5 hoặc x = - (2,5+\(\sqrt{1,25}\))
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(=x.\left(x+2\right)+1.\left(x+2\right)+x.\left(x+3\right)+1.\left(x+3\right)+x.\left(x+4\right)+1.\left(x+4\right)\)
\(=x^2+2x+x+2+x^2+3x+x+3+x^2+4x+x+4\)
\(=3x^2+12x+9\)
Vậy GTNN là 9
Dùng công thức tính tổng
\(1+2+3+...+x=11325\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=11325\)
\(\Leftrightarrow x\left(x+1\right)=22650\)
\(\Leftrightarrow x\left(x+1\right)=150.151\)
Nên x = 150
Vậy ,,,
\(1+2+3+4+...+x=11325\)
\(\Rightarrow\frac{x\left(x+1\right)}{2}=11325\)
\(\Rightarrow x\left(x+1\right)=11325\times2\)
\(\Rightarrow x\left(x+1\right)=22650\)
\(\Rightarrow150\times151=22650\)
\(\Rightarrow x=150\)
a) \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(A=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)
b) \(A=-3\Rightarrow\dfrac{1}{x+5}=-3\)
\(\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{1}{3}-5=\dfrac{-16}{3}\)
\(9x^2-42x+49=\left(3x-7\right)^2=\left(3.\dfrac{-16}{3}-7\right)^2=\left(-23\right)^2=529\) \(\left(x=\dfrac{-16}{3}\right)\)
ĐKXĐ: x<>2
\(\dfrac{x-2}{8}=\dfrac{-2}{2-x}\cdot\dfrac{1}{3}\)
=>\(\dfrac{x-2}{8}=\dfrac{2}{x-2}\cdot\dfrac{1}{3}=\dfrac{2}{3\left(x-2\right)}\)
=>\(3\cdot\left(x-2\right)^2=16\)
=>\(\left(x-2\right)^2=\dfrac{16}{3}\)
=>\(\left[{}\begin{matrix}x-2=\dfrac{4}{\sqrt{3}}\\x-2=-\dfrac{4}{\sqrt{3}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\dfrac{4}{\sqrt{3}}=2+\dfrac{4\sqrt{3}}{3}=\dfrac{6+4\sqrt{3}}{3}\\x=2-\dfrac{4}{\sqrt{3}}=\dfrac{6-4\sqrt{3}}{3}\end{matrix}\right.\)