Tính
a) \(\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)
b) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)
Tính
a) \(\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)
b) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)
\(a,\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)
\(=\dfrac{9}{3\left(\sqrt{7}-4\right)}+\dfrac{\left(\sqrt{7}-4\right)\left(\sqrt{7}+4\right)}{3\left(\sqrt{7}-4\right)}\)
\(=\dfrac{9+7-16}{3\left(\sqrt{7}-4\right)}\)
\(=0\)
\(b,\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)
\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\right]\cdot2\sqrt{3}\)
\(=\left(\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\right)\cdot2\sqrt{3}\)
\(=\left(\sqrt{2}+\sqrt{3}-\sqrt{2}\right)\cdot2\sqrt{3}\)
\(=\sqrt{3}\cdot2\sqrt{3}\)
\(=6\)
#\(Toru\)
Tính
a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)
b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)
a) \(\dfrac{1}{3\sqrt{2}-2\sqrt{3}}-\dfrac{1}{2\sqrt{3}+3\sqrt{2}}\)
\(=\dfrac{1}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}-\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)
\(=\dfrac{2\sqrt{3}}{3}\)
b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)
\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)
\(=2-\sqrt{5}-2\)
\(=-\sqrt{5}\)
f)\(\dfrac{x^2-6x+9}{x^2-8x+15}=\)
l)\(\dfrac{5yx+5x+3+3y}{10xy-15x-9-6y}=\)
\(f,\dfrac{x^2-6x+9}{x^2-8x+15}\\ =\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\\ =\dfrac{x-3}{x-5}\\ l,\dfrac{5xy+5x+3+3y}{10xy-15x-9+6y}\\ =\dfrac{5x\left(y+1\right)+3\left(y+1\right)}{5x\left(2y-3\right)+3\left(2y-3\right)}\\ =\dfrac{\left(y+1\right)\left(5x+3\right)}{\left(2y-3\right)\left(5y+3\right)}\\ =\dfrac{y+1}{2y-3}\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2}{\sqrt{x}^3-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\\ =\left(\dfrac{x+2}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{2}{x+\sqrt{x}+1}\)
e)\(\dfrac{2xy-x^2+z^2-y^2}{x^2+2-y^2+2xz}=\)
g)\(\dfrac{x^3+2x^2-x-z}{x^3-3x+2}=\)
c) \(\dfrac{y^4-1}{y^3+y^2+y+1}=\)
d)\(\dfrac{2x^2-9x+7}{-2x^2-x+28}=\)
c) \(\dfrac{y^4-1}{y^3+y^2+y+1}\)
\(=\dfrac{\left(y^2+1\right)\left(y^2-1\right)}{y^2\left(y+1\right)+\left(y+1\right)}\)
\(=\dfrac{\left(y^2+1\right)\left(y+1\right)\left(y-1\right)}{\left(y+1\right)\left(y^2+1\right)}\)
\(=y-1\)
d) \(\dfrac{2x^2-9x+7}{-2x^2-x+28}\)
\(=\dfrac{2x^2-2x-7x+7}{-\left(2x^2+8x-7x-28\right)}\)
\(=\dfrac{2x\left(x-1\right)-7\left(x-1\right)}{-\left(2x-7\right)\left(x+4\right)}\)
\(=-\dfrac{\left(2x-7\right)\left(x-1\right)}{\left(2x-7\right)\left(x+4\right)}\)
\(=\dfrac{1-x}{x+4}\)
1) rút gọn
a) \(\dfrac{x^2+3x-y^2-3y}{x^2-y^2}=\)
b) \(\dfrac{x^3+3x^2-2}{x^3+3x+4}=\)
\(b,\dfrac{x^3+3x^2-2}{x^3+3x+4}=\dfrac{x^3+x^2+2x^2+2x-2x-2}{x^3+x^2-x^2-x+4x+4}\\ =\dfrac{x^2\left(x+1\right)+2x\left(x+1\right)-2\left(x+1\right)}{x^2\left(x+1\right)-x\left(x+1\right)+4\left(x+1\right)}\\ =\dfrac{\left(x+1\right)\left(x^2+2x-2\right)}{\left(x+1\right)\left(x^2-x+4\right)}=\dfrac{x^2+2x-2}{x^2-x+4}\)
\(a,\dfrac{x^2+3x-y^2-3y}{x^2-y^2}=\dfrac{\left(x^2-y^2\right)+\left(3x-3y\right)}{x^2-y^2}\\ =\dfrac{\left(x-y\right)\left(x+y\right)+3\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\\ =\dfrac{\left(x-y\right)\left(x+y+3\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+3}{x+y}\)
Bài 1: Cho tam giác cân ABC (AB=AC). BD và CE là hai phân giác của gam giác:
a,CM:tam giác ABD=tam giác AEC.
b,CM: BCDE là hình thang cân.
GIÚP MÌNH VỚI PLSSSS
a: BD là phân giác của góc ABC
=>\(\widehat{ABD}=\dfrac{1}{2}\cdot\widehat{ABC}\left(1\right)\)
CE là phân giác của góc ACB
=>\(\widehat{ACE}=\dfrac{1}{2}\cdot\widehat{ACB}\left(2\right)\)
ΔABC cân tại A
=>\(\widehat{ABC}=\widehat{ACB}\left(3\right)\)
Từ (1),(2),(3) suy ra \(\widehat{ABD}=\widehat{ACE}\)
Xét ΔABD và ΔACE có
\(\widehat{ABD}=\widehat{ACE}\)
AB=AC
\(\widehat{BAD}\) chung
Do đó: ΔABD=ΔACE
b: ΔABD=ΔACE
=>AD=AE
Xét ΔABC có \(\dfrac{AD}{AC}=\dfrac{AE}{AB}\)
nên DE//BC
Xét tứ giác BEDC có DE//BC
nên BEDC là hình thang
Hình thang BEDC có \(\widehat{EBC}=\widehat{DCB}\)
nên BEDC là hình thang cân
|x - 1/2| <= 1/2 - |1/4 - 1/2|
\(\left|x-\dfrac{1}{2}\right|< =\dfrac{1}{2}-\left|\dfrac{1}{4}-\dfrac{1}{2}\right|\)
=>\(\left|x-\dfrac{1}{2}\right|< =\dfrac{1}{2}-\left|\dfrac{1}{4}\right|=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\)
=>\(-\dfrac{1}{4}< =x-\dfrac{1}{2}< =\dfrac{1}{4}\)
=>\(\dfrac{1}{2}-\dfrac{1}{4}< =x< =\dfrac{1}{4}+\dfrac{1}{2}\)
=>\(\dfrac{1}{4}< =x< =\dfrac{3}{4}\)
cho ( O;R ) đường kính BC . Lấy A nằm trên ( O;R ) và AB = R
tính góc A ; B ; C và AC của tam giác ABC theo R
Xét (O) có
ΔABC nội tiếp
BC là đường kính
Do đó: ΔABC vuông tại A
Xét ΔABC vuông tại A có \(BC^2=AB^2+AC^2\)
=>\(AC^2=\left(2R\right)^2-R^2=3R^2\)
=>\(AC=R\sqrt{3}\)
Xét ΔABC vuông tại A có
\(sinC=\dfrac{AB}{BC}=\dfrac{1}{2}\)
nên \(\widehat{C}=30^0\)
ΔABC vuông tại A
=>\(\widehat{B}+\widehat{C}=90^0\)
=>\(\widehat{B}=90^0-30^0=60^0\)