Giúp em câu 1b ạ
Giúp em câu 1b và câu 4 với ạ ,pls!!!!
1.
b, \(B=\dfrac{8+2\sqrt{2}}{3-\sqrt{2}}-\dfrac{2+3\sqrt{2}}{\sqrt{2}}+\dfrac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\dfrac{2\left(2+\sqrt{2}\right)\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-\dfrac{\sqrt{2}\left(\sqrt{2}+3\right)}{\sqrt{2}}+\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)}{1-\sqrt{2}}\)
\(=4+2\sqrt{2}-\sqrt{2}-3-2-\sqrt{2}\)
\(=-1\)
Bài 1:
b: Ta có: \(B=\dfrac{8+2\sqrt{2}}{3-\sqrt{2}}-\dfrac{2+3\sqrt{2}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{2}-1}\)
\(=2\sqrt{2}\left(\sqrt{2}+1\right)-\sqrt{2}-3-2+\sqrt{2}\)
\(=4+2\sqrt{2}-5\)
\(=2\sqrt{2}-1\)
4.
a, ĐK: \(x\ge0;x\ne1\)
\(Q=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)
\(=\left[\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)
\(=\dfrac{2\sqrt{x}}{\left(1-x\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(1-x\right)^2}{2}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-x\)
b, \(Q=\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Vậy \(0\le x< 1\)
c, \(Q=\sqrt{x}-x\)
\(=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(\Rightarrow maxQ=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}\)
Giúp em câu 1b và bài 4 với ạ
1B:
a: Ta có: \(N=\sqrt{8}+\sqrt{32}+\sqrt{108}-\sqrt{27}\)
\(=2\sqrt{2}+4\sqrt{2}+6\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{2}+3\sqrt{3}\)
b: Ta có: \(M=\dfrac{2}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}\)
\(=4-2\sqrt{3}-2-\sqrt{3}\)
\(=2-3\sqrt{3}\)
Giúp mk với ạ ! Câu 1b
b)\(\left\{{}\begin{matrix}x+y=-1+m\left(1\right)\\2x-y=2m\end{matrix}\right.\)
\(\Rightarrow3x=-1+3m\)
\(\Leftrightarrow x=\dfrac{-1+3m}{3}\)
Thay \(x=\dfrac{-1+3m}{3}\) vào (1) có:
\(\dfrac{-1+3m}{3}+y=-1+m\)\(\Leftrightarrow y=-1+m-\dfrac{-1+3m}{3}=-\dfrac{2}{3}\)
Suy ra với mọi m hệ luôn có nghiệm duy nhất \(\left(x;y\right)=\left(\dfrac{-1+3m}{3};-\dfrac{2}{3}\right)\)
\(xy=\left(\dfrac{-1+3m}{3}\right).\left(-\dfrac{2}{3}\right)=10\)
\(\Leftrightarrow m=-\dfrac{44}{3}\)
Vậy...
\(\left\{{}\begin{matrix}x+y=m-1\\2x-y=2m\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}2x+2y=2m-2\\2x-y=2m\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}3y=-2\\x=m-1-y\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=\dfrac{-2}{3}\\x=m-\dfrac{1}{3}\end{matrix}\right.\)
Ta có :
\(x.y=10\text{⇔}\left(m-\dfrac{1}{3}\right).\dfrac{-2}{3}=10\)
\(\text{⇔}m=\dfrac{-44}{3}\)
a) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{x-\sqrt{x}}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{x+2\sqrt{x}+1}{x-1}\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{1}{x+\sqrt{x}}\)
Giúp mình câu 1b với ạ
\(\left(\dfrac{1}{a^2+a}-\dfrac{1}{a+1}\right):\dfrac{1-a}{a^2+2a+1}=\left(\dfrac{1}{a\left(a+1\right)}-\dfrac{1}{a+1}\right);\dfrac{1-a}{\left(a+1\right)^2}=\left(\dfrac{1}{a\left(a+1\right)}-\dfrac{a}{a\left(a+1\right)}\right):\dfrac{1-a}{\left(a+1\right)^2}=\left(\dfrac{1-a}{a\left(a+1\right)}\right).\dfrac{\left(a+1\right)^2}{1-a}=\dfrac{a+1}{a}\)
Câu 1 :
a, A = \(=3\sqrt{8}-\sqrt{8}=2\sqrt{8}\)
b, đk a khác 0 ; a khác -1 ; 1
\(B=\left(\dfrac{1-a}{a^2+a}\right):\dfrac{1-a}{a^2+2a+1}=\dfrac{a+1}{a}\)
Câu 2 :
(d) đi qua A(2;7) <=> \(2m+n=7\left(1\right)\)
(d) đi qua B(1;3) <=> \(m+n=3\left(2\right)\)
Từ (1) ; (2) => \(\left\{{}\begin{matrix}2m+n=7\\m+n=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=4\\n=-1\end{matrix}\right.\)
Vậy ptđt (d) có dạng 4x - 1 = y
Giúp em bài 1b ạ
\(f\left(x\right)+g\left(x\right)=x^2-3x+2\)
\(\Leftrightarrow x^3-2x^2+x-1+g\left(x\right)=x^2-3x+2\)
\(\Leftrightarrow g\left(x\right)=-x^3+3x^2-4x+3\)
Giúp mình câu 1b với bài 2 với ạ, cần gấp lắm, cảm ơn ạ
1b)
Song song => (d): x-y +a =0
Vì d đi qua C(2;-2) => 2- (-2)+a=0
<=>a=4
=> d: x-y+4=0
Giúp em bài 1b vs 1C vs ạ
c: \(=5\cdot\dfrac{1}{5}-3\cdot\dfrac{1}{3}=0\)
\(\text{b)}\left(\dfrac{3}{5}\right)^2-\left[\dfrac{1}{3}:3-\sqrt{16}.\left(\dfrac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\dfrac{9}{25}-\left[\dfrac{1}{9}-1\right]-1\)
\(=\dfrac{9}{25}-\left(\dfrac{-8}{9}\right)-1\)
\(=\dfrac{281}{225}-1\)
\(=\dfrac{56}{225}\)
\(\text{c)}5\sqrt{\dfrac{1}{25}}-3\sqrt{\dfrac{1}{9}}\)
\(=5.\dfrac{1}{5}-3.\dfrac{1}{3}\)
\(=1-1\)
\(=0\)
M.n giúp mk nha mk Câu 1b vs Câu 2b nha
MK ĐAG CẦN GẤP Ạ
cảm ưn nhìu
a: Xét (O) có
ΔABC nội tiếp đường tròn
AB là đường kính
Do đó: ΔABC vuông tại C
1B,2B,3A,3B,4A mọi người giúp em làm bài này với ạ