Rút gọn : A= \(\frac{\frac{\sqrt{2+\sqrt{3}}}{2}}{\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\sqrt{3x-2}=x+1\) ( ĐK: \(x\ge\frac{2}{3}\))
\(\Leftrightarrow\left(\sqrt{3x-2}\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow3x-2=x^2+2x+1\)
\(\Leftrightarrow x^2-x+3=0\)
\(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\forall x\)
mà \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)
\(\Rightarrow\)\(S=\varnothing\)
b) Ta có: \(\sqrt{x^2-2x+1}=x-1\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=x-1\)
\(\Leftrightarrow\left|x-1\right|=x-1\)
+ Với \(x< 1\)\(\Rightarrow\)\(\left|x-1\right|=1-x\)
Ta có: \(1-x=x-1\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\left(L\right)\)
+ Với \(x\ge1\)\(\Rightarrow\)\(\left|x-1\right|=x-1\)
Ta có: \(x-1=x-1\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow\)\(x\inℝ\)
Vậy \(S=ℝ\)
c) Ta có: \(\sqrt{2x+1}=x-2\) ( ĐK: \(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(\sqrt{2x+1}\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow2x+1=x^2-4x+4\)
\(\Leftrightarrow x^2-6x+3=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)-6=0\)
\(\Leftrightarrow\left(x-3\right)^2=6\)
\(\Leftrightarrow x-3=\pm\sqrt{6}\)
+ \(x-3=\sqrt{6}\)\(\Leftrightarrow\)\(x=3+\sqrt{6}\approx5,45\)\(\left(TM\right)\)
+ \(x-3=-\sqrt{6}\)\(\Leftrightarrow\)\(x=3-\sqrt{6}\approx0,55\)\(\left(TM\right)\)
Vậy \(S=\left\{5,45;0,55\right\}\)
d) Ta có: \(\sqrt{x^2-3}=x^2-3\) ( ĐK: \(x\ge\pm\sqrt{3}\))
\(\Leftrightarrow\sqrt{x^2-3}-\left(\sqrt{x^2-3}\right)^2=0\)
\(\Leftrightarrow\sqrt{x^2-3}.\left(1-\sqrt{x^2-3}\right)=0\)
+ \(\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(x^2-3=0\)\(\Leftrightarrow\)\(x^2=3\)\(\Leftrightarrow\)\(x=\pm\sqrt{3}\)\(\left(TM\right)\)
+ \(1-\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(\sqrt{x^2-3}=1\)
\(\Leftrightarrow\)\(x^2-3=1\)
\(\Leftrightarrow\)\(x^2=4\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\left(TM\right)\\x=-2\left(L\right)\end{cases}}\)
Vậy \(S=\left\{-\sqrt{3};\sqrt{3};2\right\}\)
e) Ta có: \(\sqrt{x^2-6x+9}=6-x\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)
\(\Leftrightarrow\left|x-3\right|=6-x\)
+ Với \(x< 3\)\(\Leftrightarrow\)\(\left|x-3\right|=3-x\)
Ta có: \(3-x=6-x\)
\(\Leftrightarrow0x=3\)( vô nghiệm )
+ Với \(x\ge3\)\(\Leftrightarrow\)\(\left|x-3\right|=x-3\)
Ta có: \(x-3=6-x\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{9}{2}\)\(\left(TM\right)\)
Vậy \(S=\left\{\frac{9}{2}\right\}\)
g) Ta có: \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{x-1}-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}-1\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
+ Với \(x< 2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1< 0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\)
Ta có: \(1-\sqrt{x-1}=\sqrt{x-1}-1\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)\(\left(L\right)\)
+ Với \(x\ge2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1\ge0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Ta có: \(\sqrt{x-1}-1=\sqrt{x-1}-1\)
\(\Leftrightarrow0x=0\)( vô số nghiệm )
Vậy \(S=ℝ\)
a) Ta có: \(F=\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}\ge\sqrt{1}=1\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy Min(F) = 1 khi x=2
b) \(D=\sqrt{2x^2-4x+10}=\sqrt{2\left(x-1\right)^2+8}\ge\sqrt{8}=2\sqrt{2}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(Min\left(D\right)=2\sqrt{2}\Leftrightarrow x=1\)
c) \(G=\sqrt{2x^2-6x+5}=\sqrt{2\left(x-\frac{3}{2}\right)^2+\frac{1}{2}}\ge\sqrt{\frac{1}{2}}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-\frac{3}{2}\right)^2=0\Rightarrow x=\frac{3}{2}\)
Vậy \(Min\left(G\right)=\frac{\sqrt{2}}{2}\Leftrightarrow x=\frac{3}{2}\)
đề bài đúng không z? theo tôi đề là \(\sqrt{x+2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)?!
ĐKXĐ:...
Áp dụng BĐT AM-GM:
\(\left(\sqrt{x+2}+\sqrt{6-x}\right)^2\le2\left(x+2+6-x\right)=16\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{6-x}\le4\)
Lại có \(x^2-8x+24=\left(x-4\right)^2+8\ge8\forall x\)
Vậy pt vô nghiệm.
Bài làm:
Ta có: \(P=\frac{4}{a}+\frac{4}{b}+3a+3b-2\)
\(P=\left(\frac{4}{a}+a\right)+\left(\frac{4}{b}+b\right)+2\left(a+b\right)-2\)
Áp dụng bất đẳng thức Cauchy ta được:
\(P\ge2\sqrt{\frac{4}{a}.a}+2\sqrt{\frac{4}{b}.b}+2.4-2\)
\(=4+4+8-2=14\)
Dấu "=" xảy ra khi: \(a=b=2\)
Vậy Min(P) = 14 khi a=b=2
Bài làm:
đk: \(\hept{\begin{cases}x\ge0\\y\ge0\end{cases}}\)
Ta thấy: \(\hept{\begin{cases}\sqrt{x}\ge0\left(\forall x\right)\\\sqrt{y}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\sqrt{x}+\sqrt{y}\ge0\left(\forall x,y\right)\Rightarrow\frac{\sqrt{x}+\sqrt{y}}{4}\ge0\left(\forall x,y\right)\)
=> \(A\ge0\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\sqrt{x}=0\\\sqrt{y}=0\end{cases}}\Rightarrow x=y=0\)
Vậy Min(A) = 0 khi x=y=0
a) MgCO3+2HCl - MgCl2+CO2+H2O
nMgCO3= 21/81=0,25 mol
Theo p/trình cứ
1 mol MgCO3 - 2 mol HCl - 1 mol MgCl2
0,25 mol - 0,5 mol - 0,5 mol
b) VHCl= 0,5/2=0,25l
c) mMgCl2= 0,5*95=47,5g
nZn = 6,5/ 65 = 0,1 mol
mHCI = 100. 14,6% = 14,6 (g)
nHCI = 14,6/36,5 = 0,4 (mol)
a ) Theo PTHH nH2 = nZn = 0,1 mol
\(\Rightarrow\)VH2 (điều kiện đạt tiêu chuẩn) = 0,1 . 22,4 = 2,24 (l)
b ) Theo PTHH : nZnCL2 = nZn = 0,1 mol
\(\Rightarrow\)mZnCL2 = 0,1 . 136 = 13,6 (g)
\(\Rightarrow\)m chất sau pứ = mHCI dư + mZnCI2 = 7,3 + 13,6 = 20,9 (g) mH2 = 0,1 . 2 = 0,2 (g)
Áp dụng ĐLBTKL ta có :
mdd ZnCI2 = mZn + mddHCI - mH2 = 6,5 + 100- 0,2 = 106,3 (g)
C%ddZnCI2 = 20,9/ 106,3. 100% = 19,7%
trần huy nhật, Phạm Mai Anh: trmúa hmề =))))))))))))))))))))))))))))))))))))))))))
giải được chết liền
what the hell
??????????????