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Bài 1.
a) x2 +10x =24
<=>x2 + 10x -24=0
<=>x2 -2x + 12x - 24= 0
<=>x(x-2) +12(x-2)=0
<=>(x+12)(x-2)=0
<=> \(\left[\begin{array}{nghiempt}x-2=0\\x+12=0\end{array}\right.\)
⇔ \(\left[\begin{array}{nghiempt}x=2\\x=-12\end{array}\right.\)
b)4x2 + 4x = 24
⇔ 4x2 + 4x − 24 = 0
⇔ 4 (x2 + x − 6) = 0
⇔ x2 + x − 6 = 0
⇔ x2 + 3x − 2x − 6 = 0
⇔ x (x + 3) − 2 (x + 3) = 0
⇔ (x − 2)(x + 3) = 0
⇔ \(\left[\begin{array}{nghiempt}x-2=0\\x+3=0\end{array}\right.\) ⇔ \(\left[\begin{array}{nghiempt}x=2\\x=-3\end{array}\right.\)
c)4x2 − 4x = 48
⇔ 4x2 − 4x − 48 = 0
⇔ 4 (x2 − x − 12) = 0
⇔ x2 − x − 12 = 0
⇔ x2 + 3x − 4x − 12 = 0
⇔ x (x + 3) − 4 (x + 3) = 0
⇔ (x + 3)(x − 4) = 0
⇔ \(\left[\begin{array}{nghiempt}x+3=0\\x-4=0\end{array}\right.\)
<=>\(\left[\begin{array}{nghiempt}x=-3\\x=4\end{array}\right.\)
a) Mình không hiểu đề cho lắm 
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2+3\left(4x+1\right)\left(4x-1\right)\)
\(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3\)
\(=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4\)
\(=-77\)
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x\left(x^2-16\right)-x^4+1=x^3-x^4-16x+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-y^4+4=y^4-81-y^2+4=-77\)
\(\left(x+1\right)\left(x^2-x-x^2+x-1\right)=-\left(x+1\right)\)
\(\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2=-4a^2\)
\(\left(a^2+b^2+c^2+a^2-b^2-c^2\right)\left(a^2+b^2+c^2-a^2+b^2+c^2\right)=2a^2\left(2b^2+2c^2\right)=4a^2b^2+4a^2c^2\)
\(\left(a-5\right)^2\left(a+5\right)^2=\left(a^2-25\right)^2\)
\(\left(3a^3+1\right)^2-9a^2-\left(3a^3+1\right)^2=-9a^2\)
bài 1.
a.\(A=x^2-2xy+y^2+x^2+2xy+y^2=2\left(x^2+y^2\right)\)
b.\(B=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)=4xy\)
c.\(C=4a^2+4ab+b^2-\left(4a^2-4ab+b^2\right)=8ab\)
d.\(D=4x^2-4x+1-2\left(4x^2-12x+9\right)+4=-4x^2+20x-13\)
.bài 2
a.\(A=x^2+6x+9+x^2-9-2\left(x^2-2x-8\right)=10x+16;x=-\frac{1}{2}\Rightarrow A=9\)
b.\(B=9x^2+24x+16-x^2+16-10x=8x^2+14x+32\Rightarrow x=-\frac{1}{10}\Rightarrow B=\frac{767}{25}\)
c.\(C=x^2+2x+1-\left(4x^2-4x+1\right)+3\left(x^2-4\right)=6x-12\Rightarrow x=1\Rightarrow C=-6\)
d.\(D=x^2-9+x^2-4x+4-2x^2+8x=4x-5\Rightarrow x=-1\Rightarrow A=-9\)
Trả lời:
Bài 1: Rút gọn biểu thức:
a) A = ( x - y )2 + ( x + y )2
= x2 - 2xy + y2 + x2 + 2xy + y2
= 2x2 + 2y2
b) B = ( x + y )2 - ( x - y )2
= x2 + 2xy + y2 - ( x2 - 2xy + y2 )
= x2 + 2xy + y2 - x2 + 2xy - y2
= 4xy
c) C = ( 2a + b )2 - ( 2a - b )2
= 4a2 + 4ab + b2 - ( 4a2 - 4ab + b2 )
= 4a2 + 4ab + b2 - 4a2 + 4ab - b2
= 8ab
d) D = ( 2x - 1 )2 - 2 ( 2x - 3 )2 + 4
= 4x2 - 4x + 1 - 2 ( 4x2 - 12x + 9 ) + 4
= 4x2 - 4x + 1 - 8x2 + 24x - 18 + 4
= - 4x2 + 20x - 13
Bài 2: Rút gọn rồi tính giá trị biểu thức:
a) A = ( x + 3 )2 + ( x - 3 )( x + 3 ) - 2 ( x + 2 )( x - 4 )
= x2 + 6x + 9 + x2 - 9 - 2 ( x2 - 2x - 8 )
= 2x2 + 6x - 2x2 + 4x + 16
= 10x + 16
Thay x = 1/2 vào A, ta có:
\(A=10.\left(-\frac{1}{2}\right)+16=-5+16=11\)
b) B = ( 3x + 4 )2 - ( x - 4 )( x + 4 ) - 10x
= 9x2 + 24x + 16 - x2 + 16 - 10x
= 8x2 + 14x + 32
Thay x = - 1/10 vào B, ta có:
\(B=8.\left(-\frac{1}{10}\right)^2+14.\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)
c) C = ( x + 1 )2 - ( 2x - 1 )2 + 3 ( x - 2 )( x + 2 )
= x2 + 2x + 1 - 4x2 + 4x - 1 + 3 ( x2 - 4 )
= - 3x2 + 6x + 3x2 - 12
= 6x - 12
Thay x = 1 vào C, ta có:
\(C=6.1-12=-6\)
d) D = ( x - 3 )( x + 3 ) + ( x - 2 )2 - 2x ( x - 4 )
= x2 - 9 + x2 - 4x + 4 - 2x2 + 8x
= 4x - 5
Thay x = - 1 vào D, ta có:
\(D=4.\left(-1\right)-5=-9\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1+x^3-3x\left(x^2-1\right)\)
\(=3x^3+6x-3x^3+3x\)
\(=3x\)
d) \(100^2-99^2+98^2-97^2+...+2^2-1\)
\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+..+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+..+2+1\)
\(=\frac{\left(100+1\right)\cdot100}{2}=5050\)