\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Bào quan riboxom trong chất tế bào có chức năng gì? 

a) \(=\left(x^2+x\right)+\left(4x+4\right)=x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x+4\right)\)

b) \(=\left(x^2+2x\right)-\left(3x+6\right)=x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(x-3\right)\)

c) \(=\left(x^2-2x\right)-\left(3x-6\right)=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

d) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left[\left(x^2+5x\right)-\left(2x+10\right)\right]=3\left[x\left(x+5\right)-2\left(x+5\right)\right]=3\left(x-2\right)\left(x+5\right)\)

e) \(=-\left(3x^2-5x-2\right)=-\left[\left(3x^2-6x\right)+\left(x-2\right)\right]=-\left[3x\left(x-2\right)+\left(x-2\right)\right]=-\left(3x+1\right)\left(x-2\right)\)

f) \(x^2-7x+6=\left(x^2-x\right)-\left(6x-6\right)=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)

h) \(=4\left(x^2-9x+14\right)=4\left[\left(x^2-7x\right)-\left(2x-14\right)\right]=4\left[x\left(x-7\right)-2\left(x-7\right)\right]=4\left(x-2\right)\left(x-7\right)\)

i) \(=3\left(3x^2-8x+5\right)=3\left[\left(3x^2-3x\right)-\left(5x-5\right)\right]=3\left[3x\left(x-1\right)-5\left(x-1\right)\right]=3\left(x-1\right)\left(3x-5\right)\)

k) \(=-\left(2x^2+5x+2\right)=-\left[\left(2x^2+4x\right)+\left(x+2\right)\right]=-\left[2x\left(x+2\right)+\left(x+2\right)\right]=-\left(x+2\right)\left(2x+1\right)\)

l) \(=\left(x^2-5xy\right)-\left(2xy-10y^2\right)=x\left(x-5y\right)-2y\left(x-5y\right)=\left(x-5y\right)\left(x-2y\right)\)

m) \(=\left(x^2-2xy\right)-\left(xy-2y^2\right)=x\left(x-2y\right)-y\left(x-2y\right)=\left(x-2y\right)\left(x-y\right)\)

n) \(=\left(x^2-3xy\right)+\left(xy-3y^2\right)=x\left(x-3y\right)+y\left(x-3y\right)=\left(x+y\right)\left(x-3y\right)\)

Bài 2: Chọn C

Bài 4: 

a: \(\widehat{C}=180^0-80^0-50^0=50^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}< \widehat{B}\)

nên BC=AB<AC

b: Xét ΔABC có AB<BC<AC

nên \(\widehat{C}< \widehat{A}< \widehat{B}\)

Bài 2.8
Cửa hàng lãi:\(\left(1000000:800000\right)-100\%=25\%\left(giavon\right)\)
Bài 2.9
Số tiền phải bán để lãi 25% giá vốn:\(740000+\left(740000\times25\%\right)==925000\left(đồng\right)\)

ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-4;-5}

Hướng làm:

Thấy cả tử mẫu cộng lại đều bằng 2021 → Cộng thêm 1 rồi quy đồng với mỗi phân thức

\(\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\\ \Leftrightarrow\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\\ \Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}\right)=0\\ \Leftrightarrow x+2021=0\Leftrightarrow x=-2021\)

\(< =>\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\)

\(< =>\dfrac{x+2+2019}{2019}+\dfrac{x+3+2018}{2018}=\dfrac{x+4+2017}{2017}+\dfrac{x+2021}{2021}\)

\(< =>\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\)

\(< =>\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}=\right)=0\)

\(< =>x+2021=0< =>x=-2021\)

Vậy....

\(\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x}{2021}\)

\(\Leftrightarrow\frac{x+2}{2019}+1+\frac{x+3}{2018}+1=\frac{x+4}{2017}+1+\frac{x}{2021}+1\)

\(\Leftrightarrow\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2021}\)

\(\Leftrightarrow x+2021=0\)

\(\Leftrightarrow x=-2021\)

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2SO_4}=0,1.2,5=0,25\left(mol\right)\)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     0,2       0,2             0,2          0,2

Ta có: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\) ⇒ Zn hết, H₂SO₄ dư

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c) \(m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)

  \(m_{H_2SO_4\left(dư\right)}=\left(0,25-0,2\right).98=4,9\left(g\right)\)

Bài 2 : 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

100ml = 0,1l

\(n_{H2SO4}=2,5.0,1=0,25\left(mol\right)\)

a) Pt : \(Zn+2H_2SO_4\rightarrow ZnSO_4+H_2|\)

            1           1             1           1

          0,2        0,25        0,2         0,2

b) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,25}{2}\)

                     ⇒ Zn phản ứng hết , H₂SO₄ dư

                    ⇒ Tính toán dựa vào số mol của Zn

\(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,25-0,2=0,05\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

 Chúc bạn học tốt

Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a. PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

b. Đổi 100ml = 0,1 lít

Ta có: \(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{0,1}=2,5M\)

=> \(n_{H_2SO_4}=0,25\left(mol\right)\)

Ta thấy: \(\dfrac{0,2}{1}< \dfrac{0,25}{1}\)

=> H₂SO₄ dư

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

=> \(V_{H_2}=0,2.22,4=4,48\left(lít\right)\)

c. Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,2\left(mol\right)\)

=> \(m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)

Ta có: \(m_{H_2}=0,2.2=0,4\left(g\right)\)

\(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

Ta có: 

- Khối lượng ở vế trái là: 24,5 + 13 = 37,5(g)

- Khối lượng vế phải là: 0,4 + 32,2 = 32,6(g)

=> Khối lượng chất còn lại sau phản ứng là:

37,5 - 32,6 = 4,9(g)

a, Vì D,M là trung điểm AB,AC nên DM là đtb tg ABC

Do đó \(DM=\dfrac{1}{2}BC=\dfrac{7}{2}\left(cm\right)\) và DM//BC