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\(31,8^2-2.31,8.21,8+21,8^2\)
\(=\left(31,8-21,8\right)^2=10^2=100\)
\(58,2^2+2.58,2.41,8+41,8^2\)
\(=\left(58,2+41,8\right)^2=100^2=10000\)
bai 3
a) 36-12x+x2
= x2-12x+36
=(x-6)2
b) 4x2+12x+9
=(2x+3)2
c) -25x6-y8+10x3y4
=-(25x6+y8-10x3y4)
=-(5x3)2+(y4)2-10x3y4
=-(5x3)2-2.5x3y4+(y4)
=-(5x3-y4)2
d) \(\dfrac{1}{4}\) x2-5xy+25y2
=(\(\dfrac{1}{2}\) x-5y)2
Bài3:
Bạn =kia làm r nhé
Bài4:
\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =50.100=5000\\ b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\\ c,31,8^2-2.31,8.21,8+21,8^2\\ =\left(31,8-21,8\right)^2\\ =10^2=100\\ d,58,2+2.58,2.41,8+41,8^2\\ =\left(58,2+41,8\right)^2\\ =100^2=1000\)
\(4x^3-36x=0\)
\(x.\left[\left(2x\right)^2-6^2\right]=0\)
\(x.\left(2x-6\right)\left(2x+6\right)=0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\2x-6=0\end{cases}}\)hoặc \(2x+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)hoặc \(x=-3\)
KL:...............................................
Bài 1 : Tìm x, biết :
\(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\) \(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+\left(x-2\right)\left(2\left(x+2\right)-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2\left(x+2\right)-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\\left(x+2\right)^2+2>0\end{matrix}\right.\Rightarrow x=2\)
\(x\left(x-1\right)-3x+3=0\)
<=> \(x\left(x-1\right)-3\left(x-1\right)=0\)
<=> \(\left(x-3\right)\left(x-1\right)=0\)
<=> \(\hept{\begin{cases}x-3=0\\x-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=3\\x=1\end{cases}}\)
\(3x\left(x-2\right)+10-5x=0\)
<=> \(3x\left(x-2\right)+5\left(2-x\right)=0\)
<=> \(3x\left(x-2\right)-5\left(x-2\right)=0\)
<=> \(\left(3x-5\right)\left(x-2\right)=0\)
<=> \(\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)
học tốt
Bài 1:
a: \(\Leftrightarrow4x\left(x^2-9\right)=0\)
=>x(x-3)(x+3)=0
hay \(x\in\left\{0;3;-3\right\}\)
b: \(\Leftrightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)
=>(2x-6)(4x-4)=0
=>x=1 hoặc x=3
c: \(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)
=>(-2x-4)(12x-4)=0
=>x=1/3 hoặc x=-2
Bài 1:
a) \(\left(n+2\right)^2-\left(n-2\right)^2=n^2+4n+4-\left(n^2-4n+4\right)=8n\) \(⋮\)\(8\) (đpcm)
b) \(\left(n+7\right)^2-\left(n-5\right)^2=n^2+14n+49-\left(n^2-10n+25\right)=24n-24\)\(⋮\)\(24\) (đpcm)
Bài 2:
mk biến đổi về pt tích sau đó bạn giải nốt nhé
a) \(\left(x-4\right)^2-36=0\)
<=> \(\left(x-4-6\right)\left(x-4+6\right)=0\)
<=> \(\left(x-10\right)\left(x+2\right)=0\)
................
b) \(4x^2-12x=-9\)
<=> \(4x^2-12x+9=0\)
<=> \(\left(2x-3\right)^2=0\)
..............
c) \(\left(x+8\right)^2=121\)
<=> \(\left(x+8\right)^2-121=0\)
<=> \(\left(x+8+11\right)\left(x+8-11\right)=0\)
<=> \(\left(x+19\right)\left(x-3\right)=0\)
...................
Bài 3:
a) \(31,8^2-2\times31,8\times21,8+21,8^2\)
\(=\left(31,8-21,8\right)^2=10^2=100\)
b) mạo phép chỉnh đề
\(58,2^2+2\times58,2\times41,8+41,8^2\)
\(=\left(58,2+41,8\right)^2=100^2=10000\)