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+) <=> \(x^3-3x^2+3x-1+3x^2+6x+8-x^3=17\)
<=>9x=10
<=> x=\(\frac{10}{9}\)
+) \(x\left(x^2-25\right)-x^3-8=3\)<=> \(x^3-x^3-25x=3+8\)
<=> x=\(-\frac{11}{25}\)
a) \(x^3+3.2x^2y+3.2^2.x.y^2+\left(2y\right)^3=\left(x+2y\right)^3\)
b) áp dụng HDT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
c) cũng áp dụng hdt :\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2=\left[3\left(x+5\right)-x+7\right]\left[3\left(x+5\right)+x-7\right]\)\(=\left(3x+15-x+7\right)\left(2x+15+x-7\right)=\left(2x+22\right)\left(3x+8\right)=2\left(x+11\right)\left(3x+8\right)\)
d) áp dụng típ \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)=\left(x-9y\right)\left(9x-y\right)\)
e)Áp dụng típ Hdt như trên
\(\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)=\left(4y-34\right)\left(11y-22\right)\)
\(=2\left(2y-17\right).11\left(y-2\right)=22\left(2y-17\right)\left(y-2\right)\)
Bạn 1 cái t i c k nha thật sự rất cảm ơn
\(x^2+4x+3=0\)
\(x^2+x+3x+3=0\)
\(x\left(x+1\right)+3\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+3\right)=0\)
\(\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-1\\x=-3\end{array}\right.\)
\(4x^2+4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(\left[\begin{array}{nghiempt}2x-1=0\\2x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=1\\2x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{3}{2}\end{array}\right.\)
\(x^2-x-12=0\)
\(x^2-4x+3x-12=0\)
\(x\left(x-4\right)+3\left(x-4\right)=0\)
\(\left(x-4\right)\left(x+3\right)=0\)
\(\left[\begin{array}{nghiempt}x-4=0\\x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)
\(x^2-25-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+5-1\right)=0\)
\(\left(x-5\right)\left(x+4\right)=0\)
\(\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)
\(x^2\left(x^2+1\right)-x^2-1=0\)
\(x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)
\(\left(x^2+1\right)\left(x^2-1\right)=0\)
\(\left(x^2+1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+1=0\end{array}\right.\) (vì \(x^2+1\ge1>0\))
\(\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\)
\(\left(2+1\right)\left(2^2+1\right)...\left(2^8+1\right)-2^{16}=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^8+1\right)-2^{16}\)\(2^{16}\)
\(=-1\)
1. Thu gọn đa thức:
\(\left(2x-3\right)^2-\left(2x-5\right)\left(2x+5\right)\)
\(=4x^2-12x+9-4x^2+25\)
\(=-12x+34=2\left(17-6x\right)\)
2. Tìm x
\(a.\left(2x+5\right)^2-2x\left(2x-1\right)=6\left(x+1\right)\)
\(\Leftrightarrow4x^2+20x+25-4x^2+2x=6x+6\)
\(\Leftrightarrow22x+25-6x-6=0\)
\(\Leftrightarrow16x+19x=0\Leftrightarrow x=\frac{-19}{16}\)
\(b.9x^2-6x=8\)
\(\Leftrightarrow9x^2-6x-8=0\Leftrightarrow9x^2+6x-12x-8=0\)
\(\Leftrightarrow3x\left(3x+2\right)-4\left(3x+2\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\hept{\orbr{\begin{cases}3x-4=0\Rightarrow x=\frac{4}{3}\\3x+2=0\Rightarrow x=\frac{-2}{3}\end{cases}}}\)
3.CMR
\(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\)
Ta có:
\(VT=a^2+2ab+b^2+a^2-2ab+b^2\)
\(=2a^2+2b^2=2\left(a^2+b^2\right)=VP\left(đpcm\right)\)
- x2.(x3-x2+x-1)
- x.( x3-3x2-1)+3
- x.(x2-xy-y2)
Tìm x:
x3-16x = 0
=> x.(x2-16) = 0
=> x = 0 hay x2-16 = 0
=> x = 0 hay x2 = 0+16
=> x = 0 hay x2 = 16
=> x = 0 hay x = 4 hay x = -4
a, <=> (x-1)^2-4=0
<=> (x-1-2).(x-1+2)=0
<=> (x-3).(x+1)=0
<=> x-3=0 hoặc x+1=0
<=> x=3 hoặc x=-1
b, <=> x^2-x+2x-2=0
<=> x^2+x-2=0
<=> (x^2-x)+(2x-2)=0
<=> (x-1).(x+2)=0
<=> x-1=0 hoặc x+2=0
<=> x=1 hoặc x=-2
c, <=> (2x+1)^2=x^2
<=> 2x+1=x hoặc 2x+1=-x
<=> x=-1 hoặc x=-1/3
d, <=> (x^2-2x)-(3x-6)=0
<=> (x-2).(x-3)=0
<=> x-2=0 hoặc x-3=0
<=> x=2 hoặc x=3
Tk mk nha
a,\(\left(x^2-2x+1\right)-4=0\)
\(\Leftrightarrow\left(x-1\right)^2-4=0\)
\(\Leftrightarrow\left(x-1-2\right)\left(x-1+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
a, \(\left(x-15\right)\left(x+15\right)-\left(x+2\right)^2-\left(x-5\right)^2\)
\(=x^2-225-x^2-4x-4-x^2+10x-25\)
\(=-x^2+6x-254\)
b, \(\left(2x-1\right)\left(2x+1\right)+\left(x+9\right)^2-\left(x-3\right)^2\)
\(=4x^2-1+x^2+18x+81-x^2+6x-9=4x^2+24x+71\)
c, \(\left(7x-3\right)^2-\left(x-5\right)\left(x+5\right)-\left(2x+4\right)^2\)
\(=49x^2-42x+9-x^2+25-4x^2-16x-16=44x^2-58x+18\)