Mình làm gộm 2 ý luôn nhé

Ta có : \(Q\left(x\right)=5x+3x^2+5+x^2+2x^4=5x+4x^2+5+2x^4\)

Ta có : \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=\left(x^4-5x+2x^2+1\right)+\left(5x+4x^2+5+2x^4\right)\)

\(=x^4-5x+2x^2+1+5x+4x^2+5+2x^4\)

\(=5x^4+6x^2+6\)

Mà : \(5x^4+6x^2\ge0\forall x\)

Nên : \(5x^4+6x^2+6\ge6\forall x\)

Suy ra : M(x) > 0 với mọi x

Vậy M(x) vô nghiệm

a) P(x) = x⁴ - 5x + 2x² + 1 = x⁴ + 2x² - 5x + 1 

Q(x) = 5x + 3x² + 5 + 1x² + x⁴.2 = 2x⁴ + 4x² + 5x + 5

        P(x) = x⁴ + 2x² - 5x + 1
+
        Q(x) = 2x⁴ + 4x² + 5x + 5
_________________________
P(x)+Q(x) = 3x⁴ + 6x² + 6

b) Ta có: \(\hept{\begin{cases}3x^4\ge0\\6x^2\ge0\end{cases}}\forall x\)

\(\Rightarrow3x^4+6x^2\ge0\forall x\)

\(\Rightarrow M\left(x\right)=3x^4+6x^2+6\ge6>0\forall x\)

Vậy M(x) không có nghiệm

a)

P(x) +Q(x)=(X⁴-5X+2x²+1)+(5x+3x²+5+1x²+x⁴2)

=2x⁴.2 +6x² +6

\(a,\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-16\right)-\left(4x^2+20x+25\right)=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1\\ \Leftrightarrow9x^2-16-4x^2-20x-25=5x^2-6x+27\\ \Leftrightarrow5x^2-20x-41=5x^2-5x+27\\ \Leftrightarrow-15x=68\\ \Leftrightarrow x=-\dfrac{68}{15}\)Vậy..

Câu sau cũng tương tự nhé

a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)

\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)

\(\Leftrightarrow-7x+12x=20+2\)

\(\Leftrightarrow5x=22\)

\(\Rightarrow x=\dfrac{22}{5}\)

tick cho mk nha

b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)

\(x_1=3;x_2=\dfrac{-11}{10}\)

Tick cho mk nha

c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)

\(\Leftrightarrow42x-39=4\)

\(\Leftrightarrow42x=4+39\)

\(\Leftrightarrow42x=43\)

\(\Rightarrow x=\dfrac{43}{42}\)

Tick cho mk nha

c: \(P\left(-1\right)=-3-5-4+2+6+4=0\)

Vậy: x=-1 là nghiệm của P(x)

\(Q\left(-1\right)=4+1+3+2-7+1=4< >0\)

=>x=-1 không là nghiệm của Q(x)

1.

\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)

\(f\left(x\right)=0\Rightarrow x=7\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)

2.

\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)

\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)

\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)

3.

\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)

4.

\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow-6< x< 2\)

5.

\(f\left(x\right)=\frac{x^2-3x-2}{-\left(x-\frac{1}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\frac{3\pm\sqrt{17}}{2}\)

\(f\left(x\right)>0\Rightarrow\frac{3-\sqrt{17}}{2}< x< \frac{3+\sqrt{17}}{2}\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3-\sqrt{17}}{2}\\x>\frac{3+\sqrt{17}}{2}\end{matrix}\right.\)

6.

\(f\left(x\right)=\frac{\left(x-1\right)\left(x^2+x-4\right)}{\left(x-1\right)^2\left(x^2-2x-5\right)}=\frac{x^2+x-4}{\left(x-1\right)\left(x^2-2x-5\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{6}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\left\{\frac{-1\pm\sqrt{17}}{2}\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{-1-\sqrt{17}}{2}< x< 1-\sqrt{6}\\1< x< \frac{-1+\sqrt{17}}{2}\\x>1+\sqrt{6}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{-1-\sqrt{17}}{2}\\1-\sqrt{6}< x< 1\\\frac{-1+\sqrt{17}}{2}< x< 1+\sqrt{6}\end{matrix}\right.\)