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anh ơi, vậy là sai đề hả anh, chứ đề kêu chứng minh phương trình vô nghiệm mà em thấy anh ghi x=2
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Ta có: \(-16x^2-8x-3\)
\(=-\left(16x^2+8x+1+2\right)\)
\(=-\left(4x+1\right)^2-2< 0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=16x^2+8x+2=\left(16x^2+8x+1\right)+1=\left(4x+1\right)^2+1\)
Do \(\left\{{}\begin{matrix}\left(4x+1\right)^2\ge0\\1>0\end{matrix}\right.\) ;\(\forall x\)
\(\Rightarrow P=\left(4x+1\right)^2+1>0;\forall x\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=16x^2+8x+3=\left(4x\right)^2+2.4x.1+1+2\)
\(=\left(4x+1\right)^2+2>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
pt đã cho tương đương với (4x2-x+6)2=0
phần còn lại cậu tự giải đc
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\(\dfrac{8x^2+16x^2+8x}{4x^2+4x}\)
= \(\dfrac{24x^2+8x}{4x^2+4x}\)
= \(\dfrac{4x(6x+2)}{4x(x+1)}\)
= \(\dfrac{6x+2}{x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(x^2-8x+20=\left(x^2-8x+16\right)+4=\left(x-4\right)^2+4\ge4>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>(4x-1)2=0
=>4x-1=0
hay x=1/4
b: =>(x+4)(x-2)=0
=>x=-4 hoặc x=2
c: =>x2+2x+1+y2+2y+1=0
\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2=0\)
=>x=-1và y=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ:\(x\ne\pm\dfrac{1}{2}\)
\(\dfrac{1+8x}{4+8x}-\dfrac{4x}{12x-6}+\dfrac{32x^2}{3\left(4-16x^2\right)}=0\)
\(\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{4x}{6\left(2x-1\right)}+\dfrac{32x^2}{-6\cdot\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow\dfrac{6\cdot\left(1+8x\right)\left(2x-1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{4\cdot4x\left(2x+1\right)}{24\left(2x-1\right)\left(2x+1\right)}-\dfrac{32x^2\cdot4}{24\left(2x-1\right)\left(2x+1\right)}=0\)
\(\Leftrightarrow96x^2-36x-6-36x^2-16x-144x^2=0\)
\(\Leftrightarrow-84x^2-52x-6=0\)
\(\Leftrightarrow\Delta=688\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{52-\sqrt{688}}{-168}=\dfrac{-13+\sqrt{43}}{42}\\x_2=\dfrac{52+\sqrt{688}}{-168}=\dfrac{-13-\sqrt{43}}{43}\end{matrix}\right.\)
Vậy pt có 2 nghiệm phân biệt............
\(16x^2+8x+100=\left(4x\right)^2+2.4x.1+1^2+99\\ =\left(4x+1\right)^2+99>=99>0\left(DPCM\right)\)
\(16x^2+8x+100>0\)
\(\Leftrightarrow\left(4x\right)^2+2.4x.1+1+99>0\)
\(\Leftrightarrow\left(4x+1\right)^2+99>0\left(\forall x\in R\right)\)