\(2\dfrac{2}{3}-\dfrac{5}{3}x=\dfrac{7}{10}x+\dfrac{5}{6}\)

\(\Rightarrow\dfrac{8}{3}-\dfrac{5}{3}x=\dfrac{7}{10}x+\dfrac{5}{6}\)

\(\Rightarrow\dfrac{71}{30}x=\dfrac{11}{6}\)

\(\Rightarrow x=\dfrac{55}{71}\)

- Quy ước gen:

+) Gen A quy định lá chẻ

+) Gen a quy định lá nguyên

- Kiểu gen của P:

- Lá chẻ TC: AA

- Lá nguyên TC: aa

Sơ đồ lai:

P:  Lá chẻ TC  x    Lá nguyên TC

          AA                        aa

G:       A                            a 

F₁:  Aa ( 100% lá chẻ )

Cho F₁ giao phấn với nhau

P:  F₁     x    F₁

     Aa            Aa

G:  A,a          A,a

F₂: AA  :  Aa  :  Aa  :  aa  ( 3 lá chẻ, 1 lá nguyên )

`a.` Xét tam giác ACK vuông K, có:

\(\sin C=\dfrac{AK}{AC}\) \(\Leftrightarrow\sin30=\dfrac{3}{AC}\) \(\Leftrightarrow AC=6\left(cm\right)\)

\(KC=\sqrt{AC^2-AK^2}=\sqrt{6^2-3^2}=3\sqrt{3}\left(cm\right)\)

`b.`

Ta có:

\(\cot B=\dfrac{BK}{AK}\) ; \(\cot C=\dfrac{CK}{AK}\)

\(\Rightarrow\cot B+\cot C=\dfrac{BK+CK}{AK}\)

\(\Leftrightarrow\cot B+\cot C=\dfrac{BC}{AK}\)

\(\Leftrightarrow AK=\dfrac{BC}{\cot B+\cot C}\)  ( đpcm )

`c.`

Ta có: \(AK=\dfrac{BC}{\cot B+\cot C}\)

\(\Leftrightarrow AK=\dfrac{5}{\cot68+\cot30}\)

\(\Leftrightarrow AK\simeq2,34\)

\(S_{ABC}=\dfrac{1}{2}.AK.BC=\dfrac{1}{2}.2,34.5=5,86\left(cm^2\right)\)

\(x^2-y^2=21\)

\(\Rightarrow\left(x-y\right)\left(x+y\right)=21\)

\(x\left(x-1\right)+x\left(x+1\right)+\left(x-1\right)\left(x+1\right)=242\)

\(\Rightarrow x^2-x+x^2+x+x^2-1=242\)

\(\Rightarrow3x^2=243\)

\(\Rightarrow x^2=81\)

\(\Rightarrow x=\pm9\)

`@`\(A=x^2+4x+1\)

\(A=x^2+4x+4-3\)

\(A=\left(x+2\right)^2-3\ge-3\)

Dấu "=" xảy ra khi `x+2=0` 

                               `<=>x=-2`

Vậy \(Min_A=-3\) khi `x=-2`

`@`\(B=4x^2-12x-5\)

\(B=4x^2-12x+9-9-5\)

\(B=\left(2x-3\right)^2-14\ge-14\)

Dấu "=" xảy ra khi `2x-3=0`

                              `<=>x=3/2`

Vậy \(Min_B=-14\) khi `x=3/2`

`@`\(C=x^2-4x+y^2-8y+6\)

\(C=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-4-16+6\)

\(C=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y-4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Vậy \(Min_C=-14\) khi \(\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Tóm tắt: 

m₁ = 2kg       c₁ = 4200 J/kg.k            

m₂ = 0,5kg     c₂ = 880 J/kg.k

t₁ = 25^o           t₂ = 100^o

q = 44.10⁶

Q_tp = ?

m_dầu = ?

`a.` Độ tăng nhiệt độ:

\(\Delta t=t_2-t_1=100-25=75^o\)

Nhiệt lượng cung cấp cho ấm nước:

\(Q_i=Q_{nước}+Q_{ấm}=\left(m_1.c_1.\Delta t\right)+\left(m_2.c_2.\Delta t\right)\)

                              \(=\left(2.4200.75\right)+\left(0,5.880.75\right)\)

                              \(=663000\left(J\right)\)

Nhiệt lượng toàn phần:

\(Q_{tp}=\dfrac{Q_i.100}{80}=\dfrac{663000.100}{80}=828750\left(J\right)\)

\(Q_{tp}=m_{dầu}.q\Rightarrow m_{dầu}=\dfrac{Q_{tp}}{q}=\dfrac{828750}{44.10^6}=0,018\left(kg\right)\)

`a.` Thế `x=9` vào M, ta được:

\(M=\dfrac{\sqrt{9}-1}{\sqrt{9}}=\dfrac{3-1}{3}=\dfrac{2}{3}\)

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{2+8\sqrt{x}}{x-1}-\dfrac{2}{1-\sqrt{x}}\) ;\(\left(x\ge0;x\ne1\right)\)

\(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)+2+8\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\dfrac{x-3\sqrt{x}+2+2+8\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\dfrac{\left(\sqrt{x}+6\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(P=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)

\(n_X=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Đặt \(\left\{{}\begin{matrix}n_{H_2}=x\\m_{CO}=y\\n_{CO_2}=z\end{matrix}\right.\) ( mol )

\(\Rightarrow n_X=x+y+z=0,6\left(mol\right)\left(1\right)\)

\(C+H_2O\rightarrow\left(t^o\right)CO+H_2\)

\(C+2H_2O\rightarrow\left(t^o\right)CO_2+2H_2\)

\(Cu+\left\{{}\begin{matrix}CO\\H_2\end{matrix}\right.\rightarrow\left(t^o\right)CuO+\left\{{}\begin{matrix}CO_2\\H_2\end{matrix}\right.\)

\(n_{CuO}=x+y=\dfrac{40}{80}=0,5\left(mol\right)\left(2\right)\)

\(n_{H_2}=n_{CO}+2n_{CO_2}\)

\(\Rightarrow x=y+2z\left(3\right)\)

\(\left(1\right);\left(2\right);\left(3\right)\rightarrow\left\{{}\begin{matrix}x=0,35\\y=0,15\\z=0,1\end{matrix}\right.\)

\(m_X=m_{H_2}+m_{CO}+m_{CO_2}\)

       \(=0,35.2+0,15.28+0,1.44=9,3\left(g\right)\)

Bảo toàn H: \(n_{H_2O}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2O}=0,35.18=6,3\left(g\right)\)

Ta có: 9,3 gam X `->` 6,3 gam H₂O

           3,72 gam X `->` 2,52 gam H₂O

`=>` \(m=2,52\left(g\right)\)