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Bài 1:
a: Ta có: \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(k^3+128\right)\)
\(=k^3-64-k^3-128\)
=-192
b: Ta có: \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)
\(=8m^3+27n^3-27m^3+8n^3\)
\(=-19m^3+35n^3\)
Bài 4:
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x=9\)
hay x=1
b: ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x=7\)
hay \(x=\dfrac{7}{2}\)
a) Rút gọn được A = ( k 3 – 64) – (128 + k 3 ) = -192.
b) Rút gọn được B = -19 m 3 + 35 n 3 .
a) \(\left(4n^2-6nm+9m^2\right)\left(2n+3m\right)\)
\(=\left(2n+3m\right)\left[\left(2n\right)^2-2n.3m+\left(3m\right)^2\right]\)
\(=\left(2n\right)^3+\left(3m\right)^3\)
\(=8n^3+27m^3\)
b) Sửa đề \(\left(7+2b\right)\left(4b^2-14b+49\right)\)
\(=\left(7+2b\right)\left[\left(2b\right)^2-2b.7+7^2\right]\)
\(=7^3+\left(2b\right)^3\)
\(=343+8b^3\)
c) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)\)
\(=\left(5a-2b\right)\left[\left(5a\right)^2+5a.2b+\left(2b\right)^2\right]\)
\(=\left(5a\right)^3-\left(2b\right)^3\)
\(=125a^3-8b^3\)
d) \(\left(x^2+x+2\right)\left(x^2-x-2\right)\)
\(=\left[x^2+\left(x+2\right)\right]\left[x^2-\left(x+2\right)\right]\)
\(=x^4-\left(x+2\right)^2\)
a) ( 4n2 - 6mn + 9m2 ) . ( 2n + 3m ) =
= ( 2n + 3m ) . ( 4n2 - 6mn + 9m2 )
= 8n3 + 27m3
b) ( 7 + 2b ) . ( 4b2 - 14b + 49 ) =
= ( 2b + 7 ) . ( 4b2 - 14b + 49 )
= 8b3 + 343
MÌnh nghĩ là chỗ - 4b phải viết là -14b
c) ( 25a2 + 10ab + 4b2 ) . ( 5a - 2b ) =
= ( 5a - 2b ) . ( 25a2 + 10ab + 4b2 )
= 125a3 - 8b3
HOk tốt!!!!!!!!!!!!
1)\(4\left(a^4-1\right)x=5\left(a-1\right)\)
<=>x=\(\frac{5\left(a-1\right)}{a^4-1}\)
<=>x=\(\frac{5\left(a-1\right)}{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}=\frac{5}{\left(a+1\right)\left(a^2+1\right)}\)
Tương tự ta tính được y=\(\frac{4a^6+4}{5a^4-5a^2+5}\)
Suy ra x.y=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\cdot\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)
=\(\frac{5}{a+1}\)
Tương tự với x:y
\(A=\frac{4.6}{4.2}:\left(\frac{8.10}{6.8}.\frac{12.14}{10.12}.\frac{16.18}{14.16}...\frac{54.56}{54.53}\right)=\frac{6}{2}:\frac{56}{6}=\)
b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)
a ) \(\left(a^6-3a^3+9\right)\left(a^3+3\right)=a^9+27\)
b ) Đặt \(a-y=t\) , ta có :
\(\left(t-x\right)^3-\left(t+x\right)^3\)
\(=\left(t-x-t-x\right)\left[\left(t-x\right)^2+\left(t-x\right)\left(t+x\right)+\left(t+x\right)^2\right]\)
\(=-2x\left[t^2-2tx+x^2+t^2-x^2+t^2+2tx+x^2\right]\)
\(=-2x\left[\left(t^2+t^2+t^2\right)+\left(x^2-x^2+x^2\right)+\left(2tx-2tx\right)\right]\)
\(=-2x\left(3t^2+x^2\right)\)
\(=-2x\left[3\left(a-y\right)^2+x^2\right]\)
\(=-2x\left(3a^2-6ay+3y^2+x^2\right)\)
c ) \(\left(4n^2-6mn+9m^2\right)\left(2n+3m\right)=8n^3+27m^3\)
d ) \(\left(25a^2+10ab+4b^2\right)\left(5a-2b\right)=125a^3-8b^3\)
a, ( a6 - 3a3 + 9 )(a3+ 3) = (a3)3 - 33 = a9 - 27
b, ( a-x-y)3 - (a+x-y)3 = (a-x-y-a+x-y)(a-x-y+a+x-y)
= (-2y)(2a-2y) = -2y.2(a-y)
c, (4n2- 6mn + 9m2)(2n + 3m) = (2n)3 + (3m)3
= 8n3 + 27m3
d, (25a2 + 10ab +4b2)( 5a - 2b ) = 125a3 - 8b3