HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
a,Xét tam giác ABN và tam giác ACM có:góc A chungAB=AC(tam giác ABC đều)góc ANB=góc AMC(=90*)=>tam giác ABN =tam giác ACM(g-c-g)=>AN=AM(2 cạnh tương ứng)=>tam giác ANM cân tại A=>góc ANM=\(\frac{180-gócA}{2}\left(1\right)\)Có:tam giác ABC đều=>góc ACB=\(\frac{180-gócA}{2}\left(2\right)\)Từ (1) và (2)=>góc ANM =góc ACB(=\(\frac{180-gócA}{2}\))mà hai góc này ở vị trí đồng vị=>MN//BC=>NMBC là hình thangmà BN=CM(tam giác ABN=tam giác ACM)=>NMBC là hình thang cân
\(1,\left(x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)\left(x-1\right)^2\)\(2,\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)\(3,\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+2\)\(4,\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\left(\frac{1}{2}\right)^3-x^3=\frac{1}{8}-x^3\)\(5,\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x\right)^3-\left(3y\right)^3=8x^3-27y^3\)\(6,\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)
\(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)\(\Leftrightarrow\left(x^2+4x+4\right)-\left(x^2-2^2\right)=0\)\(\Leftrightarrow x^2+4x+4-x^2+4=0\)\(\Leftrightarrow4x+8=0\)
\(\Leftrightarrow4x=-8\)\(\Leftrightarrow x=-2\)\(\text{Vậy }x=-2\)
\(\left(x^2-4\right)\left(x^2-10\right)=72\)\(\Leftrightarrow x^4-10x^2-4x^2+40=72\)\(\Leftrightarrow x^4-14x^2+40-72=0\)\(\Leftrightarrow x^4-14x^2-32=0\)\(\Leftrightarrow x^4+2x^2-16x^2-32=0\)\(\Leftrightarrow x^2\left(x^2+2\right)-16\left(x^2+2\right)=0\)\(\Leftrightarrow\left(x^2+2\right)\left(x^2-16\right)=0\)\(\Leftrightarrow\left(x^2+2\right)\left(x^2-4^2\right)=0\)\(\Leftrightarrow\left(x^2+2\right)\left(x-4\right)\left(x+4\right)=0\left(1\right)\)\(Có:x^2\ge0\)\(\text{ với mọi x}\)\(\Rightarrow x^2+2\ge0+2=2\ne0\text{ với mọi x}\)\(\left(1\right)\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-4\end{array}\right.\)\(\text{Vậy }x=\pm4\)
Độ dài đáy bé của công viên là :
108 x 2/3 = 72 ( m )
Diện tích của công viên là :
( 108 + 72 ) x 62 : 2 = 5400 ( m2 )
Đổi : 5400 m2 = 54 dam2
**** cho mình nha
\(P=\left(\frac{x-1}{x+1}-\frac{x}{x-1}-\frac{3x+1}{1-x^2}\right):\frac{2x+1}{x^2}\)\(ĐKXĐ:x\ne\pm1;x\ne\frac{-1}{2}\)\(a,P=\left[\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{3x+1}{\left(x+1\right)\left(x-1\right)}\right]:\frac{2x+1}{x^2}\)\(P=\left[\frac{x^2-2x+1-x^2-x+3x+1}{\left(x+1\right)\left(x-1\right)}\right]:\frac{2x+1}{x}\)\(P=\frac{2}{\left(x+1\right)\left(x-1\right)}.\frac{x}{2x+1}\)
\(P=\frac{2x}{\left(x+1\right)\left(x-1\right)\left(2x+1\right)}\)
\(a,x^3-3x^2+1-3x\)\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)\(=\left(x^3+1^3\right)-3x\left(x+1\right)\)\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)\(=\left(x+1\right)\left(x^2-4x+1\right)\)\(c,3x^2-7x-10\)\(=3x^2+3x-10x-10\)\(=3x\left(x+1\right)-10\left(x+1\right)\)\(=\left(x+1\right)\left(3x-10\right)\)
\(\left(x^3+8y^3\right):\left(x+2y\right)\)\(=\left[x^3+\left(2y\right)^3\right]:\left(x+2y\right)\)\(=\left[\left(x+2y\right)\left(x^2-2xy+4y^2\right)\right]:\left(x+2y\right)\)\(=x^2-2xy+4y^2\)
\(\left(x+2\right)^2-\left(x-2\right)^2\)\(=\left(x+2+x-2\right)\left(x+2-x+2\right)\)\(=2x.4=8x\)
\(a,x^2+10x=24\)\(\Leftrightarrow x^2+10x-24=0\)\(\Leftrightarrow x^2-2x+12x-24=0\)\(\Leftrightarrow x\left(x-2\right)+12\left(x-2\right)=0\)\(\Leftrightarrow\left(x-2\right)\left(x+12\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)\(\text{Vậy x=2 hoặc x=-12 }\)\(b,4x^2+4x=24\)\(\Leftrightarrow4x^2+4x-24=0\)\(\Leftrightarrow4x^2-8x+12x-24=0\)\(=4x\left(x-2\right)+12\left(x-2\right)=0\)\(\Leftrightarrow\left(x-2\right)\left(4x+12\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\4x+12=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)Vậy hoặc \(\text{Vậy x=2 hoặc x=-3 }\)\(c,4x^2-4x=48\)\(\Leftrightarrow4x^2-4x-48=0\)\(\Leftrightarrow\left[\left(2x\right)^2-2.2x+1^2\right]-1^2-48=0\)\(\Leftrightarrow\left(2x-1\right)^2-49=0\)\(\Leftrightarrow\left(2x-1\right)^2-7^2=0\)\(\Leftrightarrow\left(2x-1-7\right)\left(2x-1+7\right)=0\)\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-8=0\\2x+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)\(\text{Vậy x=4 hoặc x=-3 }\)