a: Ta có: \(2x+3=x+1\)

\(\Leftrightarrow2x-x=1-3\)

hay x=-2

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

\(\Leftrightarrow4x^2-2x-4x^2-12x-9=5\)

\(\Leftrightarrow-14x=14\)

hay x=-1

c: Ta có: \(4x^2-25\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(2x-5x-10\right)\left(2x+5x+10\right)=0\)

\(\Leftrightarrow\left(-3x-10\right)\left(7x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=-\dfrac{10}{7}\end{matrix}\right.\)

d: Ta có: \(2x^2+7x+5=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e: Ta có: \(4x^2-4x=-1\)

\(\Leftrightarrow4x^2-4x+1=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

f: Ta có: \(\dfrac{1}{9}x^3-x=0\)

\(\Leftrightarrow x\left(\dfrac{1}{9}x^2-1\right)=0\)

\(\Leftrightarrow x\left(\dfrac{1}{3}x-1\right)\left(\dfrac{1}{3}x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

g: Ta có: \(x^3+3x^2+3x=7\)

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

\(\Leftrightarrow x+1=2\)

hay x=1

a, x = -2

b, x = -1

c, x = -10/3 hoặc -10/7

1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)

2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)

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

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

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

1: Ta có: \(x^3-8x+7\)

\(=x^3-x-7x+7\)

\(=x\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)

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

2: Ta có: \(x^3+8x^2-9\)

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

\(=x^2\left(x-1\right)+9\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^2+9x+9\right)\)

3: Ta có: \(3x^3-4x+1\)

\(=3x^3-3x-x+1\)

\(=3x\left(x-1\right)\left(x+1\right)-\left(x-1\right)\)

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

4: Ta có: \(x^4-3x^2+3x-1\)

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

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

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

5: Ta có: \(x^4-5x^2+4\)

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

\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

b: Ta có: \(x\left(x+1\right)-\left(2x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

d: Ta có: \(\left(x-1\right)^2-4\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x-1-2x-4\right)\left(x-1+2x+4\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)

a: \(x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

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

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

k: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

câu 2 :
\(\dfrac{-55}{132}=\dfrac{\left(-55\right):12}{132:12}=\dfrac{-5}{12}\)

câu 3:

a) <                           b) <

câu 4 :

a)\(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{5}{48}\)                 

b) \(\dfrac{-5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(\dfrac{-5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

\(\Leftrightarrow2x-1-2\sqrt{\left(2x-1\right)\left(x+3\right)}-x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=x\left(x\ge0\right)\\\sqrt{2x-1}=2\sqrt{x+3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=x^2\\2x-1=4\left(x+3\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{2}\left(loại\right)\end{matrix}\right.\)

Bài 3:

a. \(R=p\dfrac{l}{S}=1,1.10^{-6}\dfrac{30}{0,2.10^{-6}}=165\Omega\)

b. \(Q=UIt=220\left(\dfrac{220}{165}\right).15.60=254000\left(J\right)\)

Bài 2:

a. \(R=\dfrac{U^2}{P}=\dfrac{220^2}{1000}=48,4\Omega\)

b. \(Q=UIt=220\left(\dfrac{220}{48,4}\right).4.3600=14400000\left(J\right)\)

c. \(Q'=Q.40=14400000.40=576000000\left(J\right)=120000\)kWh

\(\Rightarrow T=Q'.2100=120000.2100=252000000\left(dong\right)\)

Bài 1:

a. \(Q=UIt=220.2,5.1=550\left(J\right)\)

b. \(\left\{{}\begin{matrix}Q_{thu}=mc\left(t-t_1\right)=2.4200\left(100-20\right)=672000\left(J\right)\\Q_{toa}=UIt=220.2,5.15.60=495000\left(J\right)\end{matrix}\right.\)

\(\Rightarrow H=\dfrac{Q_{toa}}{Q_{thu}}100\%=\dfrac{495000}{672000}100\%\approx73,6\%\)

2) Tìm x:

a) \(x^2+5x+6=0\)

⇒\(x^2+2x+3x+6=0\)

⇒\(\left(x^2+2x\right)+\left(3x+6\right)=0\)

⇒\(\left(x.x+2.x\right)+\left(3.x+3.2\right)=0\)

⇒\(x.\left(x+2\right)+3.\left(x+2\right)=0\)

⇒\(\left(x+2\right).\left(x+3\right)=0\)

⇒\(x+2=0\) \(hoặc\) \(x+3=0\)

\(+\))\(x+2=0\)                    \(+\))\(x+3=0\)

⇔\(x=-2\)                         ⇔\(x=-3\)

\(Vậy\) \(x\in\left\{-3;-2\right\}\)

c)\(x^2+6x+8=0\)

⇒ \(x^2+4x+2x+8=0\)

⇒ \(\left(x^2+4x\right)+\left(2x+8\right)=0\)

⇒ \(\left(x.x+4.x\right)+\left(2.x+2.4\right)=0\)

⇒ \(x.\left(x+4\right)+2.\left(x+4\right)=0\)

⇒ \(\left(x+4\right).\left(x+2\right)=0\)

⇒\(x+4=0\) \(hoặc\) \(x+2=0\)

\(+\)) \(x+4=0\)                   \(+\)) \(x+2=0\)

⇔\(x=-4\)                        ⇔\(x=-2\)

\(Vậy\) \(x\in\left\{-4;-2\right\}\)

a) x²+5x+6=0

⇒x²+2x+3x+6=0

⇒(x²+2x)+(3x+6)=0

⇒x(x+2)+3(x+2)=0

⇒(x+2)(x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

b) 9x³=x

⇒9x³-x=0

⇒9x(x²-1)=0

⇒\(\left[{}\begin{matrix}9x=0\\x^2-1=0\end{matrix}\right.\)

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

c) x²+6x+8=0

⇒ x²+2x+4x+8=0

⇒ (x²+2x)+(4x+8)=0

⇒ x(x+2)+4(x+2)=0

⇒ (x+2)(x+4)=0

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

d) x(x-3)-(x-2)²=x+1

⇒x²-3x-x²+4x-4=x+1

⇒x²-3x-x²+4x-4-x-1=0

⇒-5=0(vô lí)

e) (x+2)(x+3)-(x+1)²=0

⇒x²+5x+6-x²-2x-1=0

⇒3x+5=0

⇒3x=-5

⇒x=\(-\dfrac{5}{3}\)

f)x(x+1)-(x+1)²=0

⇒(x-x-1)(x+1)=0

⇒-1(x+1)=0

⇒x+1=0

⇒x=-1

g) (x-2)²-4(x+3)²=0

⇒x²-4x+4-4(x²+6x+9)=0

⇒x²-4x+4-4x²-24x-36=0

⇒-3x²-28x-32=0

đến đây mik chx bt lm

\(\Leftrightarrow x^2-12x+36-x^2+10x=40\)

=>-2x=4

hay x=-2