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B)(5-3x)^2=25+9x^2-30x
c)(5-x^2)(5+x^2)=25-x^4
d)(5x-1)^3=125x^3-1+15x-75x^2
e)(x^2+3)(x^4+9-3x^2)=x^6+27
f)( x-4y)(x^2+4 xy+16 y^2)= x^3-64 y^3
1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3)
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27
<=> - 3x^2 - 18x - 22 = 0
<=> 3x^2 + 18x + 22 = 0
Nửa chu vi mảnh đất là:
120 : 2 = 60 (m)
Chiều dài hơn chiều rộng là:
5 + 5 = 10 (m)
Chiều rộng là:
( 60 - 10 ) : 2 = 25 (m)
Chiều dài là:
25 + 10 = 35 (m)
Diện tích là:
25 35 = 875 ( )
\(\left(\dfrac{1}{3}.x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\left(\dfrac{1}{3}.x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)
b: \(f\left(x\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
b: \(\Leftrightarrow2\left(x^2-2x+1\right)-3x^2+5x-1=0\)
\(\Leftrightarrow2x^2-4x+2-3x^2+5x-1=0\)
\(\Leftrightarrow-x^2+x+1=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\text{Δ}=\left(-1\right)^2-4\cdot1\cdot\left(-1\right)=5\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{5}}{2}\\x_2=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
c: \(\Leftrightarrow x^2+6x+9-1-\left(x^2+8x-4x-32\right)=0\)
\(\Leftrightarrow x^2+6x+8-x^2-4x+32=0\)
=>2x+40=0
hay x=-20
d: \(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow7x^2+8x+13-7x^2+63=36\)
=>8x+76=36
hay x=-5
a)
(x3 – 3x2+3x-1) – (x3+9) +(3x2-12) = 2
3x-22 = 2
3x =24
=> x= 8
b)
x3+6x2+12x+8-6x2-x3-x2+x2 = 5
12x+8 = 5
12x = -3
=>x = -1/4
a) ( x - 1 )3 - ( x + 3 )(x2 - 3x + 9 ) + 3( x2 - 4 ) = 2
(x3-3x2\(\times\)1+3x\(\times\)12-13)-(x3+33)+(3x2-3\(\times\)4)=2
x3-3x2+3x-1-x3-9+3x2-12=2
3x-40=2
3x=42
x=14
b ) ( x + 2 )3 - 6x2 - x2 ( x + 1 ) + x2 = 5
(x3+3\(\times\)2x2+3x\(\times\)22+23)-6x2-(x3+x2)+x2=5
x3+6x2+12x+8-6x2-x3-x2+x2=5
12x+8=5
12x=\(-\)3
x=\(-\frac{1}{4}\)
bài dễ mà 
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
a) \(\left(5x-2\right)^2-\left(7-6x\right)^2=0\)
\(\Leftrightarrow\left(5x-2-7+6x\right)\left(5x-2+7-6x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}11x-9=0\\-x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{11}\\x=5\end{cases}}}\)
b) \(\left(3x-1\right)^2+\left(5x+2\right)^2=x+5\)
\(\Leftrightarrow9x^2+6x+1+25x^2+20x+4=x+5\)
\(\Leftrightarrow34x^2+26x+5=x+5\)
\(\Leftrightarrow34x^2+25x=0\)
\(\Leftrightarrow x\left(34x+25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\34x+25=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-25}{34}\end{cases}}}\)
c) Tự làm nốt
a) ( 5x - 2 )2 - ( 7 - 6x )2 = 0
<=> [ 5x - 2 - ( 7 - 6x ) ][ 5x - 2 + ( 7 - 6x ) ] = 0
<=> [ 5x - 2 - 7 + 6x ][ 5x - 2 + 7 - 6x ] = 0
<=> [ 11x - 9 ][ 5 - x ] = 0
<=> \(\orbr{\begin{cases}11x-9=0\\5-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{11}\\x=5\end{cases}}\)
b) ( 3x - 1 )2 + ( 5x + 2 )2 = x + 5
<=> 9x2 - 6x + 1 + 25x2 + 20x + 4 = x + 5
<=> 34x2 + 14x + 5 = x + 5
<=> 34x2 + 14x + 5 - x - 5 = 0
<=> 34x2 + 13x = 0
<=> 13x( 34/13x + 1 ) = 0
<=> \(\orbr{\begin{cases}13x=0\\\frac{34}{13}x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{13}{34}\end{cases}}\)
c) ( x - 2 )2 - ( 3 + 2x )2 = 20x - 4
<=> x2 - 4x + 4 - ( 4x2 + 12x + 9 ) = 20x - 4
<=> x2 - 4x + 4 - 4x2 - 12x - 9 - 20x + 4 = 0
<=> -3x2 - 36x - 1 = 0
=> Vô nghiệm ( bấm EQN ra nghiệm vô tỉ )