`|2x-3|+|x-1|=6005`

`@TH1: x < 1`

  `=>3-2x+1-x=6005`

\(<=>x=\dfrac{-6001}{3}\) (t/m)

\(@TH2: 1 \le x < \dfrac{3}{2}\)

  `=>3-2x+x-1=6005`

`<=>x=-6003` (ko t/m)

`@TH3: x \ge \dfrac{3}{2}\)

  `=>2x-3+x-1=6005`

`<=>x=2003` (t/m)

Dòng lỗi là \(TH3: x \ge \dfrac{3}{2}\) nhé.

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x-2\right)=0\)

\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) Ta có: \(a+b+c=5-3-2=0\)

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

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a:Ta có: \(x\left(x-1\right)+x=4\)

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

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

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

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

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)

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

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

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

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

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)

a)  x − 3 49 = − 2 7 ⇒ x − 3 49 = − 14 49 ⇒ x = − 11

b)  x 4 = x + 1 8 ⇒ 2 x 4 = x + 1 8 ⇒ x = 1

Giải

Bài 1

\(1,a,-40-37-\left(-29\right)\\ =-40-37+29\\ =-77+29=-48\\ b,27+\left(-36\right)-\left(-45\right)=27-36+45\\ =-9+45\\ =36\\ c,-125-\left[\left(-18\right)-125\right]\\ =-125+18+125\\ =\left(-125+125\right)+18\\ =18\\ d,140+\left[\left(-184\right)-140\right]\\ =140-184-140\\ =\left(140-140\right)-184\\ =0-184\\ =-184\)

Bài 2 

\(a,x-\left(-15\right)=-8\\ =>x+15=-8\\ =>x=-8-15\\ =>x=-23\\ b,-40-x=-35\\ =>x=\left(-40\right)-\left(-35\right)\\ =>x=-40+35\\ =>x=-5\\ c,x+\left(-50\right)=-27\\ =>x-50=-27\\ =>x=-27+50\\ =>x=23\)

1, Thực hiện phép tính:

\(a,-40+\left(-37\right)-\left(-29\right)=-77-\left(-29\right)=-48\)

\(b,27+\left(-36\right)-\left(-45\right)=-9-\left(-45\right)=36\)

\(c,-125-\left[\left(-18\right)-125\right]=-125-\left(-143\right)=18\)

\(d,140+\left[\left(-184\right)-140\right]=140+\left(-324\right)=-184\)

2, Tìm x biết:

\(a,x-\left(-15\right)=-8\)

\(x=-8+\left(-15\right)\)

\(x=-23\)

\(b,-40-x=-35\)

\(x=-40-\left(-35\right)\)

\(x=-5\)

\(c,x+\left(-50\right)=-27\)

\(x=-27-\left(-50\right)\)

\(x=23\)

 Bài 1: