(x+4)(x^2 - 4x + 16) - (x-2)(x^2+2x+4)
a) √x^2-2x+4 = 2x - 2 b) √x^2-6x+9+x = 13 c) √x^2-3x +2 = √x-1 d) √x^2-4x+4 = ✓4x^2 e) 4x^2-4x+1 = √x-8x+16
a) \(\sqrt[]{x^2-2x+4}=2x-2\)
\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)
Giải pt \(3x^2-6x=0\)
\(\Leftrightarrow3x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=2\)
c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Rút gọn các biểu thức sau
a. (x-2 ) .(x^2 +2x + 4 ) - (x^3 +2)
b. (x+4) . (x^2 -4x + 16 ) - (x-4) . (x^2 + 4x +16)
c. (x-2 )^3 - x (x+1). (x+1). (x+1) +6x( x-3)
d,(x-2).(x^2-2x+4 ) .(x+2) .(x^2 +4x +4 )
Câu d đề có đúng ko bn
mk thấy hơi sai Nguyen Thi tuong Vi
a,x^2+2x/(x+1)^2+3-x^2-2x/(x-1)^2+3=16/x^4+4x^2+16
b,x^2/x^2+2x+2+x^2/x^2-2x+2=5(x^2-5)/x^4+4+25/4
Giải phương trình:
(x+2)/(x^2+2x+4) + (x-2)/(x^2-2x+4) = 32/x(x^4+4x^2+16)
Giải phương trình:
(x+2)/(x^2+2x+4) + (x-2)/(x^2-2x+4) = 32/x(x^4+4x^2+16)
tìm x:
a.(x-3)^4-(x+3)^4+24x^3=216
b.(2x+1)(16x^4-8x^3+4x^2-2x+1)-(2x-1)(16x^4+8x^3+4x^2+2x+1)=2
tìm GTNN của bt:
x^2+2x+4
x^2-x-5/3/4
4x^2-x-3/16
giải phương trình:
\(\dfrac{x+2}{x^2+2x+4}+\)\(\dfrac{x-2}{x-2x+4}=\dfrac{32}{x\left(x^4+4x^{ }+16^{ }\right)}\)
16+x/x^2- 2x+ 18/ 2x-x^2
2y/2x^2+4x/xy_2x^2
4-x^2/ x-3+ 2x-2x^2/3-x+ 5-4x/x-3
a) \(\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
2 . ( x³ -1)-2x²(x+2x⁴) +(4x⁵+4)x=6. 3. (X²-4x+16)(x+4)-x(x+1)(x+2)+3x²= 0 4 . ( 8x +2 ) (1-3x) + ( 6x-1)(4x-10) =-50 Đề bài là tìm x nha mn Nhanh giúp mik vs
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1