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NT
4
24 tháng 6 2021
Bài 5:
1) \(\left(5+7\right)\left(7-5\right)=7^2-5^2\)
2) \(\left(x+y\right)\left(y-x\right)=y^2-x^2\)
3) \(\left(x-y\right)\left(-x-y\right)=-\left(x+y\right)\left(x-y\right)=-\left(x^2-y^2\right)=y^2-x^2\)
6) \(\left(2+3x^2\right)\left(3x^2-2\right)=9x^4-4\)
7) \(\left(\dfrac{1}{2}+x\right)\left(-x+\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+x\right)\left(\dfrac{1}{2}-x\right)=\dfrac{1}{4}-x^2\)
8) \(\left(4m-5n\right)\left(5n+4m\right)=\left(4m-5n\right)\left(4m+5n\right)=16m^2-25n^2\)
9) \(\left(7a+1\right)\left(1-7a\right)=\left(1+7a\right)\left(1-7a\right)=1-49a^2\)
10) \(\left(1+9\right)\left(1-9\right)=1-9^2\)
SV
Sinh Viên NEU
CTVVIP
10 tháng 10 2023
A
1 A
2 D
3 A
4 B
5 A
6 A
7 C
8 C
9 D
10 D
B
1 D
2 D
3 C
4 B
5 C
6 A
7 A
8 C
9 A
10 A
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 1
Hình f đề bài thiếu nên không tính được
Với hình g:
Áp dụng định lý Talet cho tam giác ADC:
\(\dfrac{AE}{ED}=\dfrac{AK}{KC}\Rightarrow\dfrac{AK}{KC}=\dfrac{4}{2}=2\)
\(\Rightarrow\dfrac{CK}{AK}=\dfrac{1}{2}\)
Áp dụng định lý Talet cho tam giác CAB:
\(\dfrac{CF}{BF}=\dfrac{CK}{AK}\Rightarrow\dfrac{x}{6}=\dfrac{1}{2}\Rightarrow x=3\)
TN
1
17 tháng 2 2022
\(\left(4A\right)\\ a,\\ \Leftrightarrow\left[\left(x-2\right)\left(2x+3\right)\right]\left[\left(x-2\right)\left(2x+3\right)\right]=0\\ \Leftrightarrow\left(-x-5\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x-5=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{-1}{3}\end{matrix}\right.\\ b,\\ \Leftrightarrow\left[3\left(2x+1\right)\right]^2-\left[2\left(x+1\right)\right]^2=0\\ \Leftrightarrow\left[3\left(2x+1\right)-2\left(x+1\right)\right]\left[3\left(2x+1\right)+2\left(x+1\right)\right]=0\\ \Leftrightarrow\left(4x+1\right)\left(8x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-5}{8}\end{matrix}\right.\\ c,\\ \Leftrightarrow\left[\left(x+1\right)+1\right]^2=0\\ \Leftrightarrow\left(x+1\right)+1=0\\ \Leftrightarrow x+2=0\Rightarrow x=-2\\ d,\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+3\right)+\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left[\left(x-1\right)\left(x+3\right)+1\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
\(\left(4B\right)\\ a,\\ \Leftrightarrow49-14x+x^2-4\left(x+25\right)^2=0\\ \Leftrightarrow49-14x+x^2-4x^2-40x-100=0\\ \Leftrightarrow3x^2-54x-51=0\\ \Leftrightarrow-3\left(x^2+18x+17\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+17\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-17\end{matrix}\right.\\ b,\\ \Leftrightarrow4x^2\left(x^2-2x+1\right)-\left(4x^2+4x+1\right)=0\\ \Leftrightarrow x^2-6x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
\(c,\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(2-x\right)=0\\ \Leftrightarrow\left(x+1\right)\left[\left(x^2-x+1\right)-\left(2-x\right)\right]=0\\ \Leftrightarrow\left(x+1\right)\left(x^1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=-1\end{matrix}\right.\\ d,\\ \Leftrightarrow\left(x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
HN
3
DK
28 tháng 6 2016
\(10x-25-x^2=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
Chúc bạn học tốt và nhớ click cho mình với nhá!
28 tháng 6 2016
= (5x-25) + (5x - x2)
= 5(x-5) + x(5-x)
= 5(x-5) - x(x-5)
= (5 - x)(x - 5)
NT
1
5: \(=4b^2-2b+\dfrac{1}{4}-\dfrac{1}{4}+a-a^2\)
\(=\left(2b\right)^2-2\cdot2b\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(a^2-a+\dfrac{1}{4}\right)\)
\(=\left(2b-\dfrac{1}{2}\right)^2-\left(a-\dfrac{1}{2}\right)^2\)
\(=\left(2b-\dfrac{1}{2}-a+\dfrac{1}{2}\right)\left(2b-\dfrac{1}{2}+a-\dfrac{1}{2}\right)\)
\(=\left(2b-a\right)\left(2b+a-1\right)\)
6:
\(=b^2-4b+4-9c^2\)
\(=\left(b-2\right)^2-9c^2\)
\(=\left(b-2-3c\right)\left(b-2+3c\right)\)
5) \(4b^2-2b+a-a^2\)
\(=\left(2b-a\right)\left(2b+a\right)-\left(2b-a\right)\)
\(=\left(2b-a\right)\left(2b+a-1\right)\)
6) \(b^2-9c^2+4+4b\)
\(=\left(b+2\right)^2-9c^2\)
\(=\left(b+3c+2\right)\left(b-3c+2\right)\)