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Bài 2:
a.
\(3x(x-4y)-\frac{12}{5}y(y-5x)=3x^2-12xy-\frac{12}{5}y^2+12xy\)
\(=3x^2-\frac{12}{5}y^2=3.4^2-\frac{12}{5}.(-5)^2=-12\)
b.
\(u=\frac{-1}{3}; v=\frac{-2}{3}\Rightarrow u+v+1=0\)
\(2u(1+u-v)-v(1-2u+v)=2u(1+u+v-2v)+v(1+u+v-3u)\)
\(=2u.(-2v)+v(-3u)=-4uv-3uv=-7uv=-7.\frac{-1}{3}.\frac{-2}{3}=\frac{-14}{9}\)
Bài 1:
\(A=x^6-(x^6-x^5)-(x^5+x^4)+(x^4-x^3)+(x^3+x^2)-(x^2+x)+1\)
\(=-x+1=-(x-1)=-(999-1)=-998\)
1/ \(4x\left(x-1\right)-x\left(4x-2\right)=5\left(x-3\right)\)
\(\Leftrightarrow4x^2-4x-4x^2+2x=5x-15\)
\(\Leftrightarrow-2x=5x-15\)
\(\Leftrightarrow7x=-15\)
\(\Leftrightarrow x=-\dfrac{15}{7}\)
Vậy ..
2/ Đặt : \(\dfrac{1}{105}=x;\dfrac{1}{651}=y\left(x,y>0\right)\)
Ta có :
\(A=2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315-651}+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3.105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(3+\dfrac{651-1}{651}\right)+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3}.\dfrac{1}{105}\right).\dfrac{1}{651}-\dfrac{1}{105}\left(4-\dfrac{1}{651}\right)-\left(\dfrac{3}{3.105.651}+\dfrac{1}{3.105.651}\right)+\dfrac{4}{105}\)
\(=\left(2+\dfrac{1}{3}x\right)y-x\left(4-y\right)-xy-\dfrac{1}{3}xy+4x\)
\(=2y+\dfrac{1}{3}xy-4x+xy-xy-\dfrac{1}{3}xy+4x\)
\(=2y\)
\(=\dfrac{2}{651}\)
Có sửa lại đề 1 chút á :>
a: Ta có: x=31
nên x-1=30
Ta có: \(A=x^3-30x^2-31x+1\)
\(=x^3-x^2\left(x-1\right)-x^2+1\)
\(=x^3-x^3+x^2-x^2+1\)
=1
c: Ta có: x=16
nên x+1=17
Ta có: \(C=x^4-17x^3+17x^2-17x+20\)
\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)
\(=20-x=4\)
d: Ta có: x=12
nên x+1=13
Ta có: \(D=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\)
\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+10\)
\(=10-x\)
=-2
d: Ta có: x=12
nên x+1=13
Ta có: \(D=x^{10}-13x^9+13x^8-13x^7+...+13x^2-13x+10\)
\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...+x^2\left(x+1\right)-x\left(x+1\right)+10\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...+x^3+x^2-x^2-x+1+9\)
\(=-x+10=-2\)
