help mình cảm ơn
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,x=16\Rightarrow A=\dfrac{16-1}{\sqrt{16}}=\dfrac{15}{4}\)
\(2,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\left(dl:x>0,x\ne1\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{x-1}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{x-1}\\ =\dfrac{4x\sqrt{x}}{x-1}\)
\(3,P=A.B=\dfrac{x-1}{\sqrt{x}}.\dfrac{4x\sqrt{x}}{x-1}=4x\)
\(\sqrt{P}>P\Leftrightarrow\sqrt{4x}>4x\Leftrightarrow\left(\sqrt{4x}\right)^2>\left(4x\right)^2\Leftrightarrow4x>16x^2\Leftrightarrow4x-16x^2>0\Leftrightarrow4x\left(1-4x\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x>0\\1-4x>0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>0\\x< \dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{x|0< x< \dfrac{1}{4}\right\}\) thì \(\sqrt{P}>P\)
\(4,\left|P\right|>P\Leftrightarrow\left|4x\right|>4x\)
\(TH_1:x\ge0\\4x>4x\Leftrightarrow4x-4x>0\Leftrightarrow0>0\left(VL\right) \)
\(TH_2:x< 0\\ -4x>4x\Leftrightarrow-4x-4x>0\Leftrightarrow-8x>0\Leftrightarrow x< 0\)
Vậy \(x< 0\) thì \(\left|P\right|>P\)
a) Từ trái qua phải
\(Na_2O+H_2O\rightarrow2NaOH\)
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)
\(Na_2SO_4+BaCl_2\rightarrow2NaCl+BaSO_4\downarrow\)
b) Từ trái qua phải
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)
\(Fe\left(OH\right)_2\xrightarrow[]{t^o}FeO+H_2O\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
\(2FeCl_2+4H_2SO_{4\left(đ,n\right)}\rightarrow Fe_2\left(SO_4\right)_3+4HCl+SO_2\uparrow+2H_2O\)
Sửa lại pt : \(FeCl_2+H_2SO_4\rightarrow FeSO_4+2HCl\)
3x . 2 + 15 = 33
3x . 2 = 33 - 15 = 18
3x = 18 : 2 = 9 = 32
=> x = 2
tham khảo tại:https://hoc24.vn/hoi-dap/tim-kiem?q=Cho+%C4%91a+th%E1%BB%A9c+f+(x)+=+ax3+bx2+cx+dax%5E3+bx%5E2+cx+d++v%E1%BB%9Bi++a+l%C3%A0+s%E1%BB%91+nguy%C3%AAn+d%C6%B0%C6%A1ng+.+Bi%E1%BA%BFt+f+(5)+-+f+(+4+)+=2012+.++Ch%E1%BB%A9ng+minh+f+(7)+-+f+(2)+l%C3%A0+h%E1%BB%A3p+s%E1%BB%91+.&id=249516
`@` `\text {Ans}`
`\downarrow`
`18,`
`2^(3x+1) = 16`
`=> 2^(3x + 1) = 2^4`
`=> 3x+1 = 4`
`=> 3x = 4 - 1`
`=> 3x = 3`
`=> x = 1`
Vậy, `x = 1`
`=> A`
`19,`
`9^12 * 27^5`
`= (3^2)^12 * (3^3)^5`
`= 3^24 * 3^15`
`=3^39`
`=> D`
`20,`
`100 < 5^(2x - 1) < 5^5`
`=> 2x - 1 < 5`
`=> x < 3 => x \in {0; 1; 2}`
`=> D`
`21,`
`5^x < 90`
`=> x < 3`
`=> x \in {0; 1; 2}`
`=> B`
`22,`
`(7x - 11)^3 = 5^2 * 2^5 + 200`
`=> (7x - 11)^3 = 25 * 32 + 200`
`=> (7x - 11)^3 = 800 + 200`
`=> (7x - 11)^3 = 1000`
`=> (7x - 11)^3 = 10^3`
`=> 7x - 11 = 10`
`=> 7x = 21`
`=> x = 3`
`=> C`
`23,`
`(x - 4)^5 = (x - 4)^4`
`=> (x - 4)^5 - (x - 4)^4 = 0`
`=> (x - 4)^4 * (x - 4 - 1) = 0`
`=> \text {TH1:} (x - 4)^4 = 0`
`=> x - 4 = 0`
`=> x = 4`
`\text {TH2:} x - 4 - 1 = 0`
`=> x - 5 = 0`
`=> x = 5`
Tổng của `2` stn x thỏa mãn dk là: `4 + 5 = 9`
`=> A.`
mình cảm ơn