\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)

\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)

\(\Leftrightarrow x=-22\)

bạn làm rõ hơn đi

mai mk học rồi , giúp nha, I x I là giá trị tuyệt đối của x !!!

\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

\(=\dfrac{-1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

a: TH1: x>=2

A=x+x-2=2x-2

TH2: x<2

A=x+2-x=2

b: TH1: x>=3

A=x-3-x=-3

TH2: x<3

A=3-x-x=-2x+3

c: TH1: x>=1

C=x-x+1=1

TH2: x<1

C=x+x-1=2x-1

d: TH1: m>=3

C=m-3-2m=-3-m

TH2: m<3

C=-m+3-2m=-3m+3

e: TH1: m>=1

E=m-m+1=1

TH2: m<1

E=m+m-1=2m-1

Đề có sai không bạn?

\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)

\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)

\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)

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

\(=\dfrac{-5+x}{5x}\)

\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)

\(=\dfrac{x+1}{x+3}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-2x+x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x-6}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)

\(=\dfrac{x^2+x-6}{x^2+x-6}\)

\(=1\)

Bài 1 

a, -25 . 63 - 25 . 3

= 25 . (-63) - 25 . 3

= 25 . [(-63) - 3]

= 25 . (-66) = -1650

Bài 2

c, (x + 1)² = 16

=> (x + 1)² = 4²

=> x + 1 = 4 

=> x = 3

d, (-38) - (x - 2) = -16

=> (x - 2) = -38 - (-16) 

=> x - 2 = -38 + 16 = -22

=> x = 2 + (-22) 

=> x = -20