\(\dfrac{S_{AOD}}{S_{DOC}}=\dfrac{10}{20}=\dfrac{1}{2}\)

=>\(AO=\dfrac{1}{2}CO\)

Vì ABCD là hình thang có hai đáy là AB và CD

và AC cắt BD tại O

nên \(S_{AOD}=S_{BOC}\)

=>\(S_{BOC}=10\left(cm^2\right)\)

\(AO=\dfrac{1}{2}OC\)

=>\(S_{AOB}=\dfrac{1}{2}\cdot S_{BOC}=5\left(cm^2\right)\)

\(S_{ABCD}=S_{ABO}+S_{BOC}+S_{DOC}+S_{AOD}\)

\(=5+10+20+10=45\left(cm^2\right)\)

a/

\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)

\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)

\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)

b/

\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)

\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)

c/

\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)

\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)

a) Ta có:

(a - b) ⋮ 6

12b ⋮ 6

⇒ [(a - b) + 12b] ⋮ 6

⇒ (a - b + 12b) ⋮ 6

⇒ (a + 11b) ⋮ 6

b) Ta có:

(a + 11b) ⋮ 6 (cmt)

12a ⋮ 6

12b ⋮ 6

⇒ [12a + 12b - (a + 11b)] ⋮ 6

⇒ (12a + 12b - a - 11b) ⋮ 6

⇒ (11a + b) ⋮ 6

E tk nha:

\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)

Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)

\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)

Vậy \(x=2\)

\(2,ĐK:x\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)

\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)

Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)

Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy ...

\(3,ĐK:x\ge-1\\ PT\Leftrightarrow3\left(x^2-x+1\right)-2\left(x+1\right)=5\sqrt{x^3+1}\) 

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(PT\Leftrightarrow3b^2-2a^2=5ab\\ \Leftrightarrow2a^2+5ab-3b^2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\a=-3b\left(vn\right)\end{matrix}\right.\Leftrightarrow a=2b\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\\x=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\left(\text{giống bài 2}\right)\)

\(\Leftrightarrow x^2-2x+1-1=0\)

\(\Leftrightarrow\left(x-1\right)^2=1\Rightarrow\left(x-1\right)=\pm1\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

a) Điện trở tương đương của đoạn mạch :

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{60.40}{60+40}=24\left(\Omega\right)\)

b) Hiệu điện thế giữa hai đầu đoạn mạch :

\(U=I.R_{tđ}=2.24=48\left(V\right)\)

⇒ \(U=U_1=U_2=48\left(V\right)\) (vì R₁ // R₂)

Cường độ dòng điện chạy qua mỗi điện trở :

\(I_1=\dfrac{U_1}{R_1}=\dfrac{48}{60}=0,8\left(A\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{48}{40}=1,2\left(A\right)\)

 Chúc bạn học tốt

\(5^{x+1}+5^{x-1}=130\)

\(5^x\cdot5^1+5^x\div5^1=130\)

\(5^x\cdot5^1+5^x\cdot\dfrac{1}{5}=130\)

\(5^x\cdot\left(5+\dfrac{1}{5}\right)=130\)

\(5^x\cdot\dfrac{26}{5}=130\)

\(5^x=130\div\dfrac{26}{5}\)

\(5^x=130\cdot\dfrac{5}{26}\)

\(5^x=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Mọi người còn câu trả lời nào khác không cứ trả lời đi mik tick cho

5^x+1.5^x-1=130

5^x-1.5²+5^x-1.1=130

5^x-1.(5²+1)=130

5^x-1.(25+1)=130

5^x-1.26=130

5^x-1=130:26

5^x-1=5¹

^{=> x-1=1}

^x=1+1

^x=2

^{Tick cho mình nha}