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c) Ta có: \(P=2x+\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{-x}{x+1}=2x+\dfrac{1}{x+1}\)
\(\Leftrightarrow\dfrac{-x}{x+1}=\dfrac{2x\left(x+1\right)+1}{x+1}\)
Suy ra: \(2x^2+2x+1=-x\)
\(\Leftrightarrow2x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: Để \(P=2x+\dfrac{1}{x+1}\) thì \(x=-\dfrac{1}{2}\)
`a)F=((x+1)/(1-x)-(1-x)/(x+1)-(4x^2)/(x^2-1)):(4x^2-4)/(x^2-2x+1)`
`đk:x ne +-1`
`F=((-(x+1)^2+(x-1)^2-4x^2)/(x^2-1)):(4(x-1)(x+1))/(x-1)^2`
`=(-x^2-2x-1+x^2-2x+1-4x^2)/(x^2-1):(4(x+1))/(x-1)`
`=(-4x^2-4x)/((x-1)(x+1)).(x-1)/(4(x+1))`
`=(-4(x-1))/((x-1)(x+1)).(x-1)/(4(x+1))`
`=-4/(x+1).(x-1)/(4(x+1)`
`=(1-x)/(x+1)^2`
`F<-1`
`<=>(1-x-(x+1)^2)/(x+1)^2<0`
Vì `(x+1)^2>0`
`=>1-x-(x+1)^2<0`
`<=>(x+1)^2+x-1>0`
`<=>x^2+2x+1+x-1>0`
`<=>x^2+3x>0`
`<=>x(x+3)>0`
`<=>` $\left[ \begin{array}{l}x>0\\x<-3\end{array} \right.$
x^3-6x^2+12x-8=0
-> x^3-2x^2-4x^2+8x+4x-8=0
-> x^2(x-2)-4x(x-2)+4(x-2)=0
-> (x-2)(x^2-4x+4)=0
->(x-2)(x-2)^2=0
-> (x-2)^3=0
->x-2=0
-> x=2 .
x^3-6x^2+12x-8=0
-> x^3-2x^2-4x^2+8x+4x-8=0
-> x^2(x-2)-4x(x-2)+4(x-2)=0
-> (x-2)(x^2-4x+4)=0
->(x-2)(x-2)^2=0
-> (x-2)^3=0
->x-2=0
-> x=2 .
nha ><
1.
\(\left(x+y\right)^2=\left(\dfrac{1}{2}.2x+\dfrac{1}{3}.3y\right)^2\le\left(\dfrac{1}{4}+\dfrac{1}{9}\right)\left(4x^2+9y^2\right)=\dfrac{169}{36}\)
\(\Rightarrow-\dfrac{13}{6}\le x+y\le\dfrac{13}{6}\)
Dấu "=" lần lượt xảy ra tại \(\left(-\dfrac{3}{2};-\dfrac{2}{3}\right)\) và \(\left(\dfrac{3}{2};\dfrac{2}{3}\right)\)
2.
\(\left(y-2x\right)^2=\left(\dfrac{1}{4}.4y+\left(-\dfrac{1}{3}\right).6x\right)^2\le\left(\dfrac{1}{16}+\dfrac{1}{9}\right)\left(16y^2+36x^2\right)=\dfrac{25}{16}\)
\(\Rightarrow\left|y-2x\right|\le\dfrac{5}{4}\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\mp\dfrac{2}{5};\pm\dfrac{9}{20}\right)\)
3.
\(B^2=\left(6.\sqrt{x-1}+8\sqrt{3-x}\right)^2\le\left(6^2+8^2\right)\left(x-1+3-x\right)=200\)
\(\Rightarrow B\le2\sqrt{10}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{x-1}}{6}=\dfrac{\sqrt{3-x}}{8}\Leftrightarrow x=\dfrac{43}{25}\)
\(B=6\sqrt{x-1}+6\sqrt{3-x}+2\sqrt{3-x}\ge6\sqrt{x-1}+6\sqrt{3-x}\)
\(B\ge6\left(\sqrt{x-1}+\sqrt{3-x}\right)\ge6\sqrt{x-1+3-x}=6\sqrt{2}\)
\(B_{min}=6\sqrt{2}\) khi \(\sqrt{3-x}=0\Rightarrow x=3\)
4.
\(49=\left(3a+4b\right)^2=\left(\sqrt{3}.\sqrt{3}a+2.2b\right)^2\le\left(3+4\right)\left(3a^2+4b^2\right)\)
\(\Rightarrow3a^2+4b^2\ge\dfrac{49}{7}=7\)
Dấu "=" xảy ra khi \(a=b=1\)
Tôi làm tạm theo cách này nhé.
\(x^4-2x^2-114x-1295\)
\(=\frac{d}{dx}\left(x^4-2x^2-114x-1295\right)\)
\(=4x^3-4x-114-0\)
\(=4x^3-4x-114\)
Bạn Phương Lê Nhật ơi!!!!
Đây là Toán 8 bạn ạ
Bạn giải mk ko hiểu j cả
Giải cụ thể đc ko bạn ạ
\(\left(x+2\right)\left(x-2\right)-x\left(x-3\right)\)
\(=x^2-4-x^2+3x=3x-4\)
Bạn tách nhỏ câu hỏi ra nhé