a, \(\Delta'=\left(m-1\right)^2-\left(-2m+5\right)=m^2-2m+1+2m-5=m^2-4\)

Để pt vô nghiệm thì \(m^2-4< 0\Leftrightarrow-2< m< 2\)

Để pt có nghiệm kép thì \(m^2-4=0\Leftrightarrow m=\pm2\)

Để pt có 2 nghiệm phân biệt thì \(m^2-4>0\Leftrightarrow\left[{}\begin{matrix}m< -2\\m>2\end{matrix}\right.\)

2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-2m+5\end{matrix}\right.\)

\(a,ĐKXĐ:x_1,x_2\ne0\\ \dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=2\\ \Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=2\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\\ \Leftrightarrow\left(2m-2\right)^2-4\left(-2m+5\right)=0\\ \Leftrightarrow4m^2-8m+4+8m-20=0\\ \Leftrightarrow4m^2-16=0\\ \Leftrightarrow m=\pm2\)

\(b,x_1+x_2+2x_1x_2\le6\\ \Leftrightarrow2m-2+2\left(-2m+5\right)\le6\\ \Leftrightarrow2m-2-4m+10-6\le0\\ \Leftrightarrow-2m+2\le0\\ \Leftrightarrow m\ge1\)

tính lỗi tí

a, `2/5 xx 3/7 +2/5 xx 4/7 = 2/5 xx (3/7+4/7)=2/5 xx 7/7 =2/5xx1=2/5`

b, `5/8xx5/4-5/8xx3/4=5/8xx(5/4-3/4)=5/8xx2/4=5/8xx1/2=5/16`

1, Ta có: \(\Delta'=\left(-m\right)^2-\left(2m-1\right)=m^2-2m+1=\left(m-1\right)^2\ge0\)

Suy ra pt luôn có 2 nghiệm

2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m-1\end{matrix}\right.\)

\(A=\left(x_1^2+x_2^2\right)-5x_1x_2\\ =\left(x_1+x_2\right)^2-7x_1x_2\\ =\left(2m\right)^2-7\left(2m-1\right)\\ =4m^2-14m+7\)

Đề sai r bạn

\(b,4m^2-14m+7\\ =4\left(m^2-\dfrac{7}{2}m+\dfrac{7}{4}\right)\\ =4\left(m^2-2.\dfrac{7}{4}m+\dfrac{49}{16}-\dfrac{21}{16}\right)\\ =4\left(m-\dfrac{7}{4}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow m=\dfrac{7}{4}\)

Vậy m=`7/4` thì A đạt GTNN

\(1,\\ a,\left\{{}\begin{matrix}4x-7y=-12\\3x+2y=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}12x-21y=-36\\12x+8y=120\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x-7y=-12\\29y=156\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x-7y=-12\\29y=156\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x-7.\dfrac{456}{29}=-12\\y=\dfrac{156}{29}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{186}{29}\\y=\dfrac{156}{29}\end{matrix}\right.\)

\(b,2x^2-5x-1=0\)

\(\Delta=\left(-5\right)^2-4.2.\left(-1\right)=25+8=33>0\)

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+\sqrt{33}}{4}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-\sqrt{33}}{4}\end{matrix}\right.\)

\(c,3x\left(x-8\right)-x=5\\ \Leftrightarrow3x^2-24x-x-5=0\\ \Leftrightarrow3x^2-25x-5=0\)

\(\Delta=\left(-25\right)^2-4.3.\left(-5\right)=625+60=685>0\)

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{25+\sqrt{685}}{6}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{25-\sqrt{685}}{6}\end{matrix}\right.\)

\(d,x^4-7x^2+12=0\\ \Leftrightarrow\left(x^4-3x^2\right)-\left(4x^2-12\right)=0\\ \Leftrightarrow x^2\left(x^2-3\right)-4\left(x^2-3\right)=0\\ \Leftrightarrow\left(x^2-3\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{3}\\x=\pm2\end{matrix}\right.\)

\(A=6+3x^3-2x+2x^2-3x^3-x^2-3x\\ =\left(3x^3-3x^3\right)+\left(2x^2-x^2\right)-\left(2x+3x\right)+6\\ =0+x^2-x+6\\ =x^2-x+6\)

Chiều cao hình thang là:
`12:2=6(m)`

Diện tích hình thang là:

`(18+12)xx6:2=90`(m²)

Đáp số: 90m²

Độ dài đường chéo thứ nhất là:
`64:(9-4)xx4=51,2(cm)`

Độ dài đường chéo thứ 2 là:

`51,2+64=115,2(cm)`

Diện tích hình thoi là:
`51,2 xx 115,2 :2 =2949,12`(cm²)

Đáp số: `2949,12`cm²

\(a,\dfrac{13}{21}+\dfrac{5}{7}=\dfrac{13}{21}+\dfrac{15}{21}=\dfrac{28}{21}\\ b,\dfrac{2}{5}:\dfrac{7}{10}=\dfrac{2}{5}\times\dfrac{10}{7}=\dfrac{2\times2\times5}{5\times7}=\dfrac{4}{7}\)