Bài 3: 

1: ĐKXĐ: \(x\ge1\)

2: ĐKXĐ: \(x\in R\)

3: ĐKXĐ: \(x\le1\)

4: ĐKXĐ: \(x>\dfrac{3}{2}\)

Bài 3:

a. \(R=R1+R2=15+30=45\Omega\)

b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)

Bài 4:

\(I1=U1:R1=6:3=2A\)

\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)

\(U=R.I=\left(3+15\right).2=36V\)

\(U2=R2.I2=15.2=30V\)

Bài 3: 

a: Gọi OK là khoảng cách từ O đến AB

Suy ra: K là trung điểm của AB

hay \(AK=BK=\dfrac{AB}{2}=\dfrac{8}{2}=4\left(cm\right)\)

Áp dụng định lí Pytago vào ΔOKA vuông tại K, ta được:

\(OA^2=OK^2+KA^2\)

hay OK=3(cm)

x^3 + x=0

x (x^2 +1) =0

Th1:

x=1

Th2:

x^2 +1 =0

x^2 = -1

=> x thuộc rỗng

Vậy x=0

x = 0 => 0^3 + 0 = 0 

Ấn nhầm, sửa tí. Chỗ Th1 đó:

x=0  (ko phải x=1)

Câu 4: ĐKXĐ: x>=1/2

Ta có: \(2\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

=>\(2x-2\sqrt{2x^2+5x-3}=1+x\cdot\sqrt{2x-1}-2x\cdot\sqrt{x+3}\)

=>\(2x-1-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}-x\cdot\sqrt{2x-1}+2x\cdot\sqrt{x+3}=0\)

=>\(2x-1-x\cdot\sqrt{2x-1}-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}+2x\cdot\sqrt{x+3}=0\)

=>\(\sqrt{2x-1}\left(\sqrt{2x-1}-x\right)-2\cdot\sqrt{x+3}\left(\sqrt{2x-1}-x\right)=0\)

=>\(\left(\sqrt{2x-1}-\sqrt{4x+12}\right)\left(\sqrt{2x-1}-x\right)=0\)

TH1: \(\sqrt{2x-1}-\sqrt{4x+12}=0\)

=>\(\sqrt{2x-1}=\sqrt{4x+12}\)

=>4x+12=2x-1

=>2x=-13

=>\(x=-\frac{13}{2}\) (loại)

TH2: \(\sqrt{2x-1}-x=0\)

=>\(\sqrt{2x-1}=x\)

=>\(\begin{cases}2x-1=x^2\\ x\ge0\end{cases}\Rightarrow\begin{cases}x^2-2x+1=0\\ x\ge0\end{cases}\Rightarrow\begin{cases}\left(x-1\right)^2=0\\ x\ge0\end{cases}\)

=>x-1=0

=>x=1(nhận)

Ta có: \(AB^2+HC^2=\left(AA'^2+A'B^2\right)+\left(A'H^2+A'C^2\right)\)

\(=\left(AA'^2+A'C^2\right)+\left(A'B^2+A'H^2\right)=AC^2+HB^2\)

Lại có: \(BC^2+HA^2=\left(BB'^2+B'C^2\right)+\left(B'H^2+B'A^2\right)\)

\(=\left(BB'^2+B'A^2\right)+\left(B'C^2+B'H^2\right)=AB^2+HC^2\)

\(\Rightarrow AB^2+HC^2=AC^2+HB^2=BC^2+HA^2\)

Bài 1: 

a, \(\)\(\)\(=>R2//\left[R4nt\left(R3//R5\right)\right]\)

\(=>Rtd=\dfrac{R2\left[R4+\dfrac{R3.R5}{R3+R5}\right]}{R2+R4+\dfrac{R3.R5}{R3+R5}}=\dfrac{1.\left[1+\dfrac{1}{1+1}\right]}{1+1+\dfrac{1}{1+1}}=0,6\left(ôm\right)\)

\(=>I=\dfrac{Uab}{Rtd}=\dfrac{10}{0,6}=\dfrac{50}{3}A=I1\)

\(=>Uab=U2345=10V=U2=U345\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{10}{1}=10A\)

\(=>I345=\dfrac{U345}{R345}=\dfrac{10}{1+\dfrac{1.1}{1+1}}=\dfrac{20}{3}A=I4=I35\)

\(=>U35=I35.R35=\dfrac{20}{3}.\dfrac{1.1}{1+1}=\dfrac{10}{3}V=U3=U5\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{\dfrac{10}{3}}{1}=\dfrac{10}{3}A,\)

\(=>I5=\dfrac{U5}{R5}=\dfrac{10}{3}A\)

b, \(I1=0,1A=Im=I2345\)

\(=>Uab=I2345.R2345=0,1.\dfrac{6\left[8+\dfrac{6.12}{6+12}\right]}{6+8+\dfrac{6.12}{6+12}}=0,4V\)

A

D

a