Số cần tìm là \(26:\dfrac{2}{3}=26\cdot\dfrac{3}{2}=39\)

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p: \(30x-3x=5\cdot54\)

=>\(27x=270\)

=>\(x=\dfrac{270}{27}=10\)

q: 3(x-2)+2(x+5)=29

=>3x-6+2x+10=29

=>5x+4=29

=>5x=25

=>\(x=\dfrac{25}{5}=5\)

t: (27-3x)(x-5)=0

=>3(9-x)(x-5)=0

=>(9-x)(x-5)=0

=>\(\left[{}\begin{matrix}9-x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)

v: x+(x+1)+...+(x+30)=1240

=>31x+(1+2+...+30)=1240

=>\(31x+30\cdot\dfrac{31}{2}=1240\)

=>\(31\left(x+15\right)=31\cdot40\)

=>x+15=40

=>x=40-15=25

s: (x+2)+(4x+4)+(7x+6)+...+(25x+18)+(28x+20)=1560

=>(x+4x+7x+...+25x+28x)+(2+4+6+...+20)=1560

=>\(x\left(1+4+...+28\right)+2\left(1+2+3+...+10\right)=1560\)

=>\(x\left[\left(\dfrac{28-1}{3}+1\right)\cdot\dfrac{\left(28+1\right)}{2}\right]+2\cdot\dfrac{10\cdot11}{2}=1560\)

=>\(x\left[10\cdot\dfrac{29}{2}\right]+10\cdot11=1560\)

=>\(145x=1560-110=1450\)

=>x=10

Chiều dài mảnh vườn thứ ba là: \(\dfrac{65}{12}-\dfrac{25}{6}=\dfrac{65}{12}-\dfrac{50}{12}=\dfrac{15}{12}=\dfrac{5}{4}\left(m\right)\)

Chiều dài mảnh vườn thứ nhất là:

\(\dfrac{65}{12}-\dfrac{15}{4}=\dfrac{65}{12}-\dfrac{45}{12}=\dfrac{20}{12}=\dfrac{5}{3}\left(m\right)\)

Chiều dài mảnh vườn thứ hai là:

\(\dfrac{25}{6}-\dfrac{5}{3}=\dfrac{25}{6}-\dfrac{10}{6}=\dfrac{15}{6}=\dfrac{5}{2}\left(m\right)\)

a: \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)

\(=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-2a^3\)

\(=6ab^2\)

b: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=x^3-8-x^3-3x^2-3x-1+3\left(x^2-1\right)\)

\(=-3x^2-3x-9+3x^2-3=-3x-12\)

a: \(2^x+2^{x+4}=544\)

=>\(2^x+2^x\cdot16=544\)

=>\(17\cdot2^x=544\)

=>\(2^x=32=2^5\)

=>x=5

b: \(4^{2x+1}+4^{2x}=80\)

=>\(4^{2x}\cdot4+4^{2x}=80\)

=>\(4^{2x}\cdot5=80\)

=>\(4^{2x}=16=4^2\)

=>2x=2

=>x=1

c: \(3^{2x+2}+3^{2x+1}=108\)

=>\(3^{2x}\cdot9+3^{2x}\cdot3=108\)

=>\(12\cdot3^{2x}=108\)

=>\(3^{2x}=9=3^2\)

=>2x=2

=>x=1

d: \(7^{x+3}-7^{x+1}=16464\)

=>\(7^x\cdot343-7^x\cdot7=16464\)

=>\(7^x\cdot336=16464\)

=>\(7^x=49=7^2\)

=>x=2

Dựng \(AH\perp CD;BK\perp CD\left(H;K\in CD\right)\)

Xét tg vuông ADH có

\(\widehat{DAH}=90^o-\widehat{D}=30^o\)

\(\Rightarrow DH=\dfrac{AD}{2}=\dfrac{4}{2}=2cm\) (trong tg vuông cạnh đối diện góc \(30^o\) băng nửa cạnh huyền)

\(\Rightarrow AH=\sqrt{AD^2-DH^2}=\sqrt{16-4}=\sqrt{12}=2\sqrt{3}cm\)

\(\Rightarrow AH=BK=2\sqrt{3}cm\) (đường cao của hình thang)

Xét tg vuông BCK có

\(\widehat{KBC}=90^o-\widehat{C}=45^o\)

=> tg BCK vuông cân tại K \(\Rightarrow CK=BK=2\sqrt{3}cm\)

\(\Rightarrow BC=\sqrt{BK^2+CK^2}=\sqrt{12+12}=2\sqrt{6}cm\)

Xét HCN ABKH có

\(AB=KH=CD-DH-CK=8-2\sqrt{3}-2\sqrt{3}=8-4\sqrt{3}=4\left(2-\sqrt{3}\right)cm\)

1/3 tuổi của Hải 4 năm trước bằng 1/4 tuổi của Hải 4 năm sau

=>\(\dfrac{1}{3}-\dfrac{1}{4}\) số tuổi của Hải năm nay là \(\dfrac{1}{4}\times4+\dfrac{1}{3}\times4=1+\dfrac{4}{3}=\dfrac{7}{3}\)

Tuổi của Hải năm nay là \(\dfrac{7}{3}:\dfrac{1}{12}=\dfrac{7}{3}\times12=28\left(tuổi\right)\)

Cảm ơn bạn nhé

1: \(x^2-25=\left(x-5\right)\left(x+5\right)\)

2: \(9x^2-\dfrac{1}{16}y^2=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2\)

\(=\left(3x-\dfrac{1}{4}y\right)\left(3x+\dfrac{1}{4}y\right)\)

3: \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

4: \(\left(2x-5\right)^2-64=\left(2x-5-8\right)\left(2x-5+8\right)\)

\(=\left(2x-13\right)\left(2x+3\right)\)

5: \(81-\left(3x+2\right)^2\)

\(=\left(9-3x-2\right)\left(9+3x+2\right)\)

\(=\left(-3x+7\right)\left(3x+11\right)\)

6: \(9\left(x-5y\right)^2-16\left(x+y\right)^2\)

\(=\left(3x-15y\right)^2-\left(4x+4y\right)^2\)

\(=\left(3x-15y-4x-4y\right)\left(3x-15y+4x+4y\right)\)

\(=\left(-x-19y\right)\left(7x-11y\right)\)

7: \(x^3-8=x^3-2^3=\left(x-2\right)\left(x^2+2x+4\right)\)

8: \(27x^3+125y^3=\left(3x\right)^3+\left(5y\right)^3\)

\(=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)

9: \(x^6+216=\left(x^2\right)^3+6^3\)

\(=\left(x^2+6\right)\left(x^4-6x^2+36\right)\)

10: \(x^2+8x+16=x^2+2\cdot x\cdot4+4^2=\left(x+4\right)^2\)

11: \(9x^2-12xy+4y^2\)

\(=\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\)

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

12: \(-25x^2y^2+10xy-1\)

\(=-\left[\left(5xy\right)^2-2\cdot5xy\cdot1+1^2\right]\)

\(=-\left(5xy-1\right)^2\)

13: \(x^3-6x^2+12x-8\)

\(=x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3\)

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

14: \(8x^3+12x^2y+6xy^2+y^3\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)

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

BM+MN+NC=BC
=>\(BC=MN+\dfrac{2}{3}MN+\dfrac{1}{2}MN=\dfrac{13}{6}MN\)

=>\(\dfrac{BM}{BC}=\dfrac{2}{3}:\dfrac{13}{6}=\dfrac{2}{3}\cdot\dfrac{6}{13}=\dfrac{12}{39}=\dfrac{4}{13}\)

=>BC=3,25BM

\(S_{ABM}=\dfrac{1}{2}\cdot MH\cdot AB=\dfrac{1}{2}\cdot12\cdot25=150\left(cm^2\right)\)

BC=3,25BM nên \(S_{ABC}=3,25\cdot S_{ABM}=3,25\cdot150=487,5\left(cm^2\right)\)