HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
\(C,\left(x-1\right)^5-\left(x-1\right)^4=0\)\(\Leftrightarrow\left(x-1\right)^4\left(x-1-1\right)=0\)\(\Leftrightarrow\left(x-1\right)^4\left(x-2\right)\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
49 con
chủ nhat truoc thu 2 la ngay 19
=>CN dau tien la:19-7-7=5
\(\frac{x+3}{2x-2}-\frac{4}{x^2-1}.\frac{x+1}{2}\)\(=\frac{x+3}{2x-2}-\left(\frac{4}{x^2-1}.\frac{x+1}{2}\right)\)
\(=\frac{x+3}{2\left(x-1\right)}-\frac{4\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{x+3}{2\left(x-1\right)}-\frac{4}{2\left(x-1\right)}\)\(=\frac{x+3-4}{2\left(x-1\right)}\)\(=\frac{x-1}{2\left(x-1\right)}\)\(=\frac{1}{2}\)
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)\(\text{Vậy }x=-1\)
\(a,\left(4x-8\right)\left(\frac{1}{2}-x\right)=0\)\(\Leftrightarrow4\left(x-2\right)\left(\frac{1}{2}-x\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\\frac{1}{2}-x=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{1}{2}\end{array}\right.\)\(b,2x^2-32=0\)\(\Leftrightarrow2\left(x^2-16\right)=0\)\(\Leftrightarrow x^2-16=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)
\(a,x^4-10x^3+25x^2=0\)\(x^2\left(x^2-10x+25\right)=0\)\(\Leftrightarrow x^2\left(x-5\right)^2=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=0\\\left(x-5\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)Vậy x=0 hoặc x=5\(b,x^3+3x^2+3x+1=0\)\(\Leftrightarrow\left(x+1\right)^3=0\)\(\Leftrightarrow x+1=0\)\(\Leftrightarrow x=-1\)Vậy x=-1
\(a,45+x^3-5x^2-9x\)\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)\(=x^2\left(x-5\right)+9\left(5-x\right)\)\(=x^2\left(x-5\right)-9\left(x-5\right)\)\(=\left(x-5\right)\left(x^2-9\right)\)\(=\left(x-5\right)\left(x^2-3^2\right)\)\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)\(c,2x^2+3x-5\)\(=2x^2-2x+5x-5\)\(=2x\left(x-1\right)+5\left(x-1\right)\)\(=\left(x-1\right)\left(2x+5\right)\)\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)\(\text{hay }x^2+10x+20=t\left(2\right)\)(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)\(=t^2-4^2+16\)\(=t^2-16+16\)\(=t^2\left(3\right)\)Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)
Có:Trong hình thang,hai góc kề một cạnh bên bù nhau(vì đó là hai góc trong cùng phía)=>Trong hình thang ABCD,hai góc A và D kề một cạnh bên AD;hai góc C và B kề một cạnh bên BC(1)(1)=>góc A+góc D=180* <=>góc D=180*-góc A <=>góc D =180*-130* <=>góc D=50*(1)=>góc C+góc B=180* <=>góc C=180*-góc B <=>góc C=180*-48* <=>góc C=132*Vậy góc D=50*;góc C=132*
\(x^4-4y^4\)
\(=\left(x^2\right)^2-\left(2y\right)^2\)\(=\left(x^2-2y\right)\left(x^2+2y\right)\)