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a: Δ=(-14)^2-4*4*12=196-192=4>0
=>Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x=\dfrac{14-2}{8}=\dfrac{12}{8}=\dfrac{3}{2}\\x=\dfrac{14+2}{8}=\dfrac{16}{8}=2\end{matrix}\right.\)
b: Δ=(-15)^2-4*3*5
=225-60=165>0
=>Phương trình có hai nghiệm pb là:
\(\left\{{}\begin{matrix}x=\dfrac{15-\sqrt{165}}{6}\\x=\dfrac{15+\sqrt{165}}{6}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) √x2 = 7 ⇔ |x| = 7
⇔ x1 = 7 và x2 = -7
b) √x2 = |-8| ⇔ √x2 = 8
⇔ |x| = 8 ⇔ x1 = 8 và x2 = -8
⇔ |x| = 3 ⇔ x1 = 3 và x2 = -3
⇔ |3x| = 12 ⇔ |x| = 4
⇔ x1 = 4 và x2 = -4
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: x>=-1
\(4x^2-2\sqrt{x+1}=x+2\)
=>\(4x^2-2\sqrt{x+1}-x-2=0\)
=>\(4x^2+3x-4x-3+1-2\sqrt{x+1}=0\)
=>\(\left(4x+3\right)\left(x-1\right)+1-\sqrt{4x+4}=0\)
=>\(\left(4x+3\right)\left(x-1\right)+\dfrac{1-4x-4}{1+\sqrt{4x+4}}=0\)
=>\(\left(4x+3\right)\left(x-1\right)-\dfrac{4x+3}{1+\sqrt{4x+4}}=0\)
=>\(\left(4x+3\right)\left(x-1-\dfrac{1}{1+\sqrt{4x+4}}\right)=0\)
=>4x+3=0
=>x=-3/4(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\left(x^2+x\right)2+3\left(x^2+x\right)+2\)
=\(\left(x^2+x\right)6+2\)
b,\(\left(x^2+x\right)2-2\left(x^2+x\right)-15\)
=\(-4\left(x^2+x\right)-15\)
c,\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
=\(\left(x^2+x+1\right)\left(x^2+x+1\right)+1-12\)
=\(\left(x^2+x+1\right)^2-11\)
d,\(\left(x^2+x\right)2+4x^2+4x-12\)
=\(x\left(x+1\right)2+2x\left(x+1\right)-12\)
=\(2x\left(x+1\right)+2x\left(x+1\right)-12\)
=\(\left(x+1\right)\left(2x+2x-12\right)\)
= \(\left(x+1\right)\left(4x-12\right)=4\left(x+1\right)\left(x-3\right)\)
e,\(\left(x^2+2x\right)2+9x^2+18x+20\)
=\(x\left(x+2\right)2+9x\left(x+2\right)+20\)
=\(2x\left(x+2\right)+9x\left(x+2\right)+20=\left(x+2\right)\left(2x+9x+20\right)\)
=\(\left(x+2\right)\left(11x+20\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 7:
\(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|\)
\(=\left|2\cdot5-1\right|=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,x^2=5\Leftrightarrow x=\pm\sqrt{5}\)
Vậy \(S=\left\{\pm\sqrt{5}\right\}\)
\(b,3x^2-12=0\Leftrightarrow3x^2=12\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)
Vậy \(S=\left\{\pm2\right\}\)
\(c,4x^2-3=-9\)
\(\Leftrightarrow4x^2=-6\)
\(\Leftrightarrow x^2=-\dfrac{3}{2}\) (loại)
Vậy pt vô nghiệm.
\(d,5x^2-3=-3\)
\(\Leftrightarrow5x^2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
a)
`x^2 =5`
`=>\(\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
b)
`3x^2 -12=0`
`<=>3x^2 =12`
`<=>x^2 =4`
\(< =>\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
c)
`4x^2 -3=-9`
`<=>4x^2 =-6`
`<=>x^2 =-3/2` (vô lí vì `x>=0AA x` )
d)
`5x^2 -3=3`
`<=>5x^2 =0`
`<=>x^2 =0`
`<=>x=0`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+b+c=1-\left(m+3\right)+m+2=0\)
\(\Rightarrow\) Phương trình luôn có 2 nghiệm: \(\left[{}\begin{matrix}x=1\\x=m+2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x_1=1\\x_2=m+2\end{matrix}\right.\)
\(\Rightarrow1-4\left(m+2\right)^2=-35\)
\(\Leftrightarrow\left(m+2\right)^2=9\Rightarrow\left[{}\begin{matrix}m=1\\m=-5\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x_1=m+2\\x_2=1\end{matrix}\right.\)
\(\Rightarrow4\left(m+2\right)^2-1=-35\Rightarrow4\left(m+2\right)^2=-34< 0\) (vô nghiệm)
Bài làm:
Ta có: \(\frac{x^2+4x^2}{\left(x+2\right)^2}=12\)
\(\Leftrightarrow5x^2=12\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2+48x+48=0\)
\(\Leftrightarrow7\left(x^2+\frac{48}{7}x+\frac{576}{49}\right)-\frac{240}{7}=0\)
\(\Leftrightarrow\left(x+\frac{24}{7}\right)^2-\left(\frac{4\sqrt{15}}{7}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{24-4\sqrt{15}}{7}\right)\left(x+\frac{24+4\sqrt{15}}{7}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4\sqrt{15}-24}{7}\\x=-\frac{24+4\sqrt{15}}{7}\end{cases}}\)
*Bổ sung : Thiếu ĐK rồi : \(x\ne-2\)