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Đề đúng nhé bạn :
\(B=2\dfrac{1}{315}.\dfrac{1}{651}-\dfrac{1}{105}.3\dfrac{650}{651}-\dfrac{4}{315.651}+\dfrac{4}{105}\)
\(\Leftrightarrow B=\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-\dfrac{3}{315}\left(3+\dfrac{650}{651}\right)-\dfrac{4}{315.651}+\dfrac{12}{315}\)
\(\Leftrightarrow B=\left(2+\dfrac{1}{315}\right).\dfrac{1}{651}-3.\dfrac{1}{315}\left(4-\dfrac{1}{651}\right)-4.\dfrac{1}{315}.\dfrac{1}{651}+12.\dfrac{1}{315}\)
Đặt \(\dfrac{1}{315}=a;\dfrac{1}{651}=b\) thay vào B , ta được :
\(B=\left(2+a\right)b-3a\left(4-b\right)-4ab+12a\)
\(\Leftrightarrow B=2b+ab-12a+3ab-4ab+12a\)
\(\Leftrightarrow B=2b+\left(ab+3ab-4ab\right)+\left(12a-12a\right)\)
\(\Leftrightarrow B=2b\)
\(\Leftrightarrow B=2.\dfrac{1}{651}\)
\(\Leftrightarrow B=\dfrac{2}{651}\)
Vậy \(B=\dfrac{2}{651}\)
:D
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)
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, ĐKXĐ: x≠±3
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\dfrac{-1}{x^2}\)
b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:
\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4
c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x2>0 (Đúng với mọi x)
Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)
