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3 tháng 9 2021

\(c,x\left(x-20\right)-x+20=0\\ \Leftrightarrow x^2-20x-x+20=0\\ \Leftrightarrow x^2-21x+20=0\\ \Leftrightarrow\left(x-20\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=20\\x=1\end{matrix}\right.\)

\(b,x^2-4+\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow x^2-4+x^2-x-6=0\\ \Leftrightarrow2x^2-x-10=0\\ \Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

a: Ta có: \(x\left(x-20\right)-x+20=0\)

\(\Leftrightarrow\left(x-20\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=1\end{matrix}\right.\)

b: Ta có: \(x^2-4+\left(x+2\right)\left(x-3\right)=0\)

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

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

1: \(\dfrac{11}{x^4y};\dfrac{3}{xy^3}\)

\(\dfrac{11}{x^4y}=\dfrac{11\cdot y^2}{x^4y^3}=\dfrac{11y^2}{x^4y^3}\)

\(\dfrac{3}{xy^3}=\dfrac{3\cdot x^3}{xy^3\cdot x^3}=\dfrac{3x^3}{x^4y^3}\)

2: \(\dfrac{2}{3x^3y^2};\dfrac{3}{4x^7y}\)

\(\dfrac{2}{3x^3y^2}=\dfrac{2\cdot4\cdot x^4}{3x^3y^2\cdot4x^4}=\dfrac{8x^4}{12x^7y^2}\)

\(\dfrac{3}{4x^7y}=\dfrac{3\cdot3\cdot y}{4x^7y\cdot3y}=\dfrac{9y}{12x^7y^2}\)

1C

2D

3A

4A

a:Ta có: \(A=-4x^2+x-1\)

\(=-4\left(x^2-\dfrac{1}{4}x+\dfrac{1}{4}\right)\)

\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)

b: Ta có: \(B=-3x^2+5x+6\)

\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

c: Ta có: \(C=-x^2+3x+4\)

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

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

a) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)

\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)

\(\Leftrightarrow x-1.5=0\)

hay x=1,5

b) Ta có: \(2x^3+3x^2+2x+3=0\)

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

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

18 tháng 9 2021

\(x^2-16x-15x\left(x-4\right)\)

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

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

\(=\left(x-4\right)\left(x^2-11x\right)=x\left(x-4\right)\left(x-11\right)\)

18 tháng 9 2021

\(=x\left(x-4\right)\left(x+4\right)-15x\left(x-4\right)\\ =x\left(x-4\right)\left(x-11\right)\)

2 tháng 8 2021

8) \(\dfrac{x+7}{3}+\dfrac{x+5}{4}=\dfrac{x+3}{5}+\dfrac{x+1}{6}\)

\(\Rightarrow\dfrac{x+7}{3}+\dfrac{x+5}{4}-\dfrac{x+3}{5}-\dfrac{x+1}{6}=0\)

\(\Rightarrow\dfrac{x+7}{3}+2+\dfrac{x+5}{4}+2-\dfrac{x+3}{5}-2-\dfrac{x+1}{6}-2=0+2+2-2-2\)

\(\Rightarrow\left(\dfrac{x+7}{3}+2\right)+\left(\dfrac{x+5}{4}+2\right)-\left(\dfrac{x+3}{5}+2\right)-\left(\dfrac{x+1}{6}+2\right)=0\)

\(\Rightarrow\left(\dfrac{x+7}{3}+\dfrac{6}{3}\right)+\left(\dfrac{x+5}{4}+\dfrac{8}{4}\right)-\left(\dfrac{x+3}{5}+\dfrac{10}{5}\right)-\left(\dfrac{x+1}{6}+\dfrac{12}{2}\right)=0\)

\(\Rightarrow\left(x+13\right)\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+13=0\\\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=0\end{matrix}\right.\)

\(x+13=0\)

\(\Rightarrow x=-13\)

\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=0\)

\(\dfrac{13}{60}=0\) (vô lí)

Vậy \(x=-13\)

9) Bạn chuyển vế rồi cộng 3 vào từng mỗi số

2 tháng 8 2021

mơn b nhìu aaaayeu

e) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)

5 tháng 8 2021

e)x3-4x+14x(x-2)=0

⇔ x(x2-4)+14x(x-2)=0

⇔ x(x-2)(x+2)+14x(x-2)=0

⇔ (x-2)(x2+2x+14x)=0

⇔ x(x-2)(x+16)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\x=-16\end{matrix}\right.\)

g)x2(x+1)-x(x+1)+x(x-1)=0

⇔ (x+1)(x2-x)+x(x-1)=0

⇔ x(x+1)(x-1)+x(x-1)=0

⇔ x(x-1)(x+2)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)