Chiều cao hình thang là:
\(\left(10+6\right)\times\dfrac{3}{4}=12\left(cm\right)\)

Diện tích hình thang là:

\(\left(10+6\right)\times12:2=96\left(cm^2\right)\)

Đáp số: 96cm²

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}\\ =1-0-0-...-0-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

\(g,\left(2x-5\right)^2-2\left(4x^2-25\right)=0\\ \Leftrightarrow\left(2x-5\right)^2-2\left(2x-5\right)\left(2x+5\right)=0\\ \Leftrightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(4x+10\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(2x-5-4x-10\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(-2x-15\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{15}{2}\end{matrix}\right.\)

\(h,\left(3-x\right)\left(4-x\right)=x^2-6x+9\\ \Leftrightarrow\left(3-x\right)\left(4-x\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(3-x\right)\left(4-x\right)-\left(3-x\right)^2\\ \Leftrightarrow\left(3-x\right)\left(4-x-3+x\right)=0\\ \Leftrightarrow3-x=0\\ \Leftrightarrow x=3\\ i,x^3+x^2-20x=0\\ \Leftrightarrow x\left(x^2+x-20\right)=0\\ \Leftrightarrow x\left[\left(x^2+5x\right)-\left(4x+20\right)\right]=0\\ \Leftrightarrow x\left[x\left(x+5\right)-4\left(x+5\right)\right]=0\\ \Leftrightarrow x\left(x+5\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=4\end{matrix}\right.\)

\(k,2x^2-9x-5=0\\ \Leftrightarrow\left(2x^2+10x\right)-\left(x+5\right)=0\\ \Leftrightarrow2x\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\dfrac{17}{8}-\dfrac{13}{25}-\dfrac{12}{25}+\dfrac{15}{8}\\ =\left(\dfrac{17}{8}+\dfrac{15}{8}\right)-\left(\dfrac{13}{25}+\dfrac{12}{25}\right)\\ =\dfrac{32}{8}-\dfrac{25}{25}\\ =4-1\\ =3\)

Bạn đăng từng bài 1 ra nhé

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=7\\x_1x_2=-6\end{matrix}\right.\)

\(E=2x_1^2x_2+2x_1x_2^2\\ =2x_1x_2\left(x_1+x_2\right)\\ =2.\left(-6\right).7\\ =-84\)

\(\left(x^2+x-1\right)\left(x^2+x+3\right)=5\\ \Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+4\right)-5=0\\ \Leftrightarrow\left(x^2+x-1\right)^2+4\left(x^2+x-1\right)-5=0\\ \Leftrightarrow\left[\left(x^2+x-1\right)^2+5\left(x^2+x-1\right)^2\right]-\left[\left(x^2+x-1\right)+5\right]=0\\ \Leftrightarrow\left(x^2+x-1\right)\left(x^2+x-1+5\right)-\left(x^2+x-1+5\right)=0\\ \Leftrightarrow\left(x^2+x-1+5\right)\left(x^2+x-1-1\right)=0\\ \Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{15}{4}=0\\\left(x^2+2x\right)-\left(x+2\right)=0\end{matrix}\right. \)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\\x\left(x+2\right)-\left(x+2\right)=0\end{matrix}\right.\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(\dfrac{2n+5}{n-3}=\dfrac{\left(2n-6\right)+11}{n-3}=\dfrac{2\left(n-3\right)+11}{n-3}=2+\dfrac{11}{n-3}\)

Để biểu thức trên là số nguyên thì \(\dfrac{11}{n-3}\) nguyên\(\Rightarrow11⋮\left(n-3\right)\)\(\Rightarrow n-3\inƯ\left(11\right)\)

Ta có bảng:

Vậy \(n\in\left\{-8;2;4;14\right\}\)

1, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=-6\end{matrix}\right.\)

\(A=\left(x_1-2x_2\right)\left(2x_1-x_2\right)\\ =2x_1^2-4x_1x_2-x_1x_2+2x_1^2\\ =2\left(x_1^2+x_2^2\right)-5x_1x_2\\ =2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\\ =2\left(-5\right)^2-4.\left(-6\right)-5.\left(-6\right)\\ =104\)

2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=5\\x_1x_2=-3\end{matrix}\right.\)

\(B=x_1^3x_2+x_1x_2^3\\ =x_1x_2\left(x_1^2+x_2^2\right)\\ =\left(-3\right)\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\\ =\left(-3\right)\left[5^2-2\left(-3\right)\right]\\ =-93\)

b, Thay x=1 vào g(x) ta có:

\(g\left(1\right)=100.1^{100}+99.1^{99}+...+2.1^2+1+1\\ =100.1+99.1+...+2.1+1+1\\ =100+99+...+2+1+1\\ =\left(100+1\right).100:2+1\\ =5051\)

c, Thay x=-1 vào h(x) ta có:

\(h\left(-1\right)=1-\left(-1\right)+\left(-1\right)^2-\left(-1\right)^3+...+\left(-1\right)^{100}\\ =1+1+1+...+1\\ =101\)