giúp e câu a,b vs ạ
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H9
HT.Phong (9A5)
CTVHS
19 tháng 8 2023
a) \(A=\left(2\sqrt{12}-\sqrt{75}+\dfrac{1}{2}\sqrt{48}\right):\sqrt{3}\)
\(A=\left(4\sqrt{3}-5\sqrt{3}+2\sqrt{3}\right):\sqrt{3}\)
\(A=\sqrt{3}:\sqrt{3}\)
\(A=1\)
b) \(B=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(B=\left|2-\sqrt{5}\right|-\left|\sqrt{5}+1\right|\)
\(B=-2+\sqrt{5}-\sqrt{5}-1\)
\(B=-3\)
c) \(C=\dfrac{3}{\sqrt{7}-2}-\dfrac{4}{3+\sqrt{7}}\)
\(C=\dfrac{3\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{4\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(C=\dfrac{3\left(\sqrt{7}+2\right)}{3}-\dfrac{4\left(3-\sqrt{7}\right)}{2}\)
\(C=\sqrt{7}+2-2\left(3-\sqrt{7}\right)\)
\(C=\sqrt{7}+2-6+2\sqrt{7}\)
\(C=3\sqrt{7}-4\)
d) \(D=3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\)
\(D=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-\dfrac{1}{4}\cdot8\sqrt{2a}\)
\(D=5\sqrt{2a}-3a\sqrt{2a}-2\sqrt{2a}\)
\(D=3\sqrt{2a}-3a\sqrt{2a}\)
e) \(E=\dfrac{3+\sqrt{3}}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)
\(E=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(E=\left(\sqrt{3}+1\right)-\dfrac{2\left(\sqrt{3}+1\right)}{2}\)
\(E=\left(\sqrt{3}+1\right)-\left(\sqrt{3}+1\right)\)
\(E=0\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2023
Lời giải:
a.
\(A=2\sqrt{\frac{12}{3}}-\sqrt{\frac{75}{3}}+\frac{1}{2}\sqrt{\frac{48}{3}}=2\sqrt{4}-\sqrt{25}+\frac{1}{2}\sqrt{16}\)
\(2.2-5+\frac{1}{2}.4=1\)
b.
\(B=|2-\sqrt{5}|-|\sqrt{5}+1|=\sqrt{5}-2-(\sqrt{5}+1)=-3\)
c.
\(C=\frac{3(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}-\frac{4(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}\)
\(=\frac{3(\sqrt{7}+2)}{7-2^2}-\frac{4(3-\sqrt{7})}{3^2-7}\)
\(=\frac{3(\sqrt{7}+2)}{3}-\frac{4(3-\sqrt{7})}{2}=\sqrt{7}+2-2(3-\sqrt{7})=-4+3\sqrt{7}\)
e.
\(E=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}}-\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\sqrt{3}+1-\frac{2(\sqrt{3}+1)}{3-1^2}=(\sqrt{3}+1)-(\sqrt{3}+1)=0\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
20 tháng 6 2023
\(a,\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\\ \Leftrightarrow48x^2-20x-12x+5-3x-48x^2-7+112x-81=0\\ \Leftrightarrow77x=83\\ \Leftrightarrow x=\dfrac{83}{77}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
20 tháng 6 2023
\(b,\left(x-4\right)\left(x-1\right)=\left(x-2\right)\left(x-3\right)\\ \Leftrightarrow x^2-4x-x+4=x^2-2x-3x+6\\ \Leftrightarrow x^2-x^2-4x-x+2x+3x=6-4\\ \Leftrightarrow0x=2\left(vô.lí\right)\)
Vậy không có x thoả mãn
26 tháng 9 2021
\(b,\left(1\right)4Al+3O_2\underrightarrow{^{to}}2Al_2O_3\\ \left(2\right)Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ \left(3\right)Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow3BaSO_4\downarrow+2AlCl_3\\ \left(4\right)AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ \left(5\right)Al\left(NO_3\right)_3+3KOH\rightarrow Al\left(OH\right)_3\downarrow+3KNO_3\\ \left(6\right)2Al\left(OH\right)_3\underrightarrow{^{to}}Al_2O_3+3H_2O\)
26 tháng 9 2021
\(d,\left(1\right)3Fe+2O_2\underrightarrow{^{to}}Fe_3O_4\\ \left(2\right)Fe_3O_4+4CO\underrightarrow{^{to}}3Fe+4CO_2\\ \left(3\right)FeO+H_2\underrightarrow{^{to}}Fe+H_2O\\ \left(4\right)Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\\ \left(5\right)2Fe\left(NO_3\right)_3+Fe\rightarrow3Fe\left(NO_3\right)_2\\ \left(6\right)Fe\left(NO_3\right)_2+2KOH\rightarrow Fe\left(OH\right)_2\downarrow+2KNO_3\\ \left(7\right)4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\)
11 tháng 12 2023
b: ĐKXĐ: x<>-3
\(\dfrac{3x+x^2}{x^2+x+1}\cdot\dfrac{3x^3-3}{x+3}\)
\(=\dfrac{x\left(x+3\right)}{x^2+x+1}\cdot\dfrac{3\left(x^3-1\right)}{x+3}\)
\(=\dfrac{3x\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=3x\left(x-1\right)\)
e: ĐKXĐ: \(x\notin\left\{4;-5\right\}\)
\(\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}\)
\(=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}\cdot\dfrac{2x-8}{\left(x+5\right)^2}\)
\(=\dfrac{2\cdot2\left(x-4\right)}{\left(x-4\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^2+4x+16}\)
H9
HT.Phong (9A5)
CTVHS
30 tháng 6 2023
a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=\left(2x+3\right)\left[\left(2x\right)^2-2x\cdot3+3^2\right]-2\left(4x^3-1\right)\)
\(=\left[\left(2x\right)^3+3^3\right]-2\left(4x^3-1\right)\)
\(=\left(8x^3+27\right)-8x^3+2\)
\(=8x^3+27-8x^3+2\)
\(=29\)
Vậy: ....
c) \(2\left(x^3+y^3\right)-3\left(x^3+y^3\right)\)
\(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3x^2-3y^2\)
\(=2\left(x^2-xy+y^2\right)\cdot1-3x^2-3y^2\)
\(=2x^2-2xy+2y^2-3x^2-3y^2\)
\(=-x^2-2xy-y^2\)
\(=-\left(x^2+2xy+y^2\right)\)
\(=-\left(x+y\right)^2\)
\(=-\left(1\right)^2=-1\)
Vậy: ...
8 tháng 5 2021
a, Ta có : \(\sin^2x+\cos^2x=1\)
\(\Rightarrow\sin x=\sqrt{1-\cos^2x}=\left|\dfrac{\sqrt{15}}{4}\right|\)
Mà \(0< x< \dfrac{\pi}{2}\)
\(\Rightarrow\sin x=\dfrac{\sqrt{15}}{4}\)
Ta lại có : \(\left\{{}\begin{matrix}\sin2x=2\sin x\cos x=\dfrac{\sqrt{15}}{8}\\\cos2x=2\cos^2x-1=-\dfrac{7}{8}\end{matrix}\right.\)
Vậy ...
c, Ta có : \(\tan2x=\dfrac{2\tan x}{1-\tan^2x}=\dfrac{4}{3}=\dfrac{\sin2x}{\cos2x}\)
- Ta có HPT : \(\left\{{}\begin{matrix}\sin^22x+\cos^22x=1\\3\sin2x-4\cos2x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sin2x=\left|\dfrac{4}{5}\right|\\\cos2x=\left|\dfrac{3}{5}\right|\end{matrix}\right.\)
Lại có : \(\pi< x< \dfrac{3}{2}\pi\)
\(\Rightarrow\left\{{}\begin{matrix}\sin2x=\dfrac{4}{5}\\\cos2x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy ...
\(a,2x^3-6x^2-2x\left(x^2-3x+2\right)\)
\(=2x^3-6x^2-2x^3+6x^2-4x\)
\(=\left(2x^3-2x^3\right)+\left(-6x^2+6x^2\right)-4x\)
\(=0+0-4x\)
\(=-4x\)
\(b,-6x^2\left(3x-1\right)+2x\left(9x^2+5x\right)\)
\(=-18x^3+6x^2+18x^3+10x^2\)
\(=\left(-18x^3+18x^3\right)+\left(6x^2+10x^2\right)\)
\(=0+16x^2\)
\(=16x^2\)
a. \(2x^3-6x^2-2x\left(x^2-3x+2\right)\\ =2x^3-6x^2-2x^3+6x^2-4x\\ =-4x\)
b. \(-6x^2\left(3x-1\right)+2x\left(9x^2+5x\right)\\ =-18x^3+6x^2+18x^3+10x^2\\ =6x^2+10x^2\\ =16x^2\)