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Bài 1:
\(\left(x-2\right)\left(2x+5\right)-2x^2-1=0\)
\(\Leftrightarrow2x^2+x-10-2x^2-1=0\)
\(\Leftrightarrow x-11=0\Leftrightarrow x=11\)
Bài 2:
\(P=\left|2-x\right|+2y^4+5\)
Ta thấy:
\(\begin{cases}\left|2-x\right|\ge0\\2y^4\ge0\end{cases}\)
\(\Rightarrow\left|2-x\right|+2y^4\ge0\)
\(\Rightarrow\left|2-x\right|+2y^4+5\ge5\)
\(\Rightarrow P\ge5\)
Dấu = khi \(\begin{cases}\left|2-x\right|=0\\2y^4=0\end{cases}\)\(\Leftrightarrow\)\(\begin{cases}x=2\\y=0\end{cases}\)
Vậy MinP=5 khi \(\begin{cases}x=2\\y=0\end{cases}\)
Bài 4:
2(2x+x2)-x2(x+2)+(x3-4x+13)
=2x2+4x-x3-2x2+x3-4x+13
=(2x2-2x2)+(4x-4x)-(-x3+x3)+13
=13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.\)
\(-17-\left(x-3\right)^2\)
Ta có: \(\left(x-3\right)^2\ge0\)với \(\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2\le0\)với \(\forall x\)
\(\Leftrightarrow17-\left(x-3\right)^2\le17\)với \(\forall x\)
Dấu '' = '' xảy ra khi:
\(\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy \(Max=-17\)khi \(x=3\)
\(2.\)
\(A=x\left(x+1\right)+\frac{3}{2}\)
\(A=x^2+x+\frac{3}{2}\)
\(A=\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)
\(\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)với \(\forall x\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)với \(\forall x\)
Vậy \(Max=\frac{5}{4}\)khi \(x=\frac{-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{x}+2\right)^2=\dfrac{1}{x^2}+2\cdot\dfrac{1}{x}\cdot2+4=\dfrac{1}{x^2}+\dfrac{4}{x}+4\)
Do đó, không có x trong dãy này