P = \(\left(\frac{4}{x}-\frac{2x-4}{x^2-x}\right).\frac{x^2-1}{x+1}\)

P = \(\left(\frac{4}{x}-\frac{2\left(x-2\right)}{x\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{x+1}\)

P = \(\frac{4\left(x-1\right)-2\left(x-2\right)}{x\left(x-1\right)}.\left(x-1\right)\)

P = \(\frac{4x-4-2x+4}{x\left(x-1\right)}.\left(x-1\right)\)

P = \(\frac{2x\left(x-1\right)}{x\left(x-1\right)}=2\)

Dòng vua Gia-va đã chinh phục được Xu-ma-tơ-ra, thống nhất Indonesia dưới Vương triều Mô-giô-pa-hít

Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (a ≠ c; b ≠ d)

CMR: \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\)

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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\dfrac{a+c}{a-c}=\dfrac{bk+dk}{bk=dk}=\dfrac{\left(b+d\right)k}{\left(b-d\right)k}=\dfrac{b+d}{b-d}\)

Mà \(\dfrac{b+d}{b-d}=\dfrac{b+d}{b-d}\)

=> \(\dfrac{a+c}{a-c}=\dfrac{b+d}{b-d}\) (đpcm)

b)\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{81}{64}\)

=> \(\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{9}{8}\right)^2\)

=> \(x+\dfrac{1}{2}=\dfrac{9}{8}\)

=> \(x=\dfrac{9}{8}-\dfrac{1}{2}\)

=> \(x=\dfrac{5}{8}\)

c) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và \(x^2-2y^2+z^2=44\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

=> \(\left[{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có:

\(x^2-2y^2+z^2=44\)

=> \(\left(2k\right)^2-2\left(3k\right)^2+\left(5k\right)^2=44\)

=> \(4k^2-2.9k^2+25k^2=44\)

=> \(4k^2-18k^2+25k^2=44\)

=> \(\left(4-18+25\right).k^2=44\)

=> \(11.k^2=44\)

=> \(k^2=4\) => \(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

*Với k = 2, ta có:

_\(x=2.2=4\)

_\(y=3.2=6\)

_\(z=5.2=10\)

*Với k = -2, ta có:

_\(x=2.\left(-2\right)=-4\)

_\(y=3.\left(-2\right)=-6\)

_\(z=5.\left(-2\right)=-10\)

Đặt \(\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{6}=t\)

=> \(\left[{}\begin{matrix}x=2k-1\\y=4k-3\\z=6k-5\end{matrix}\right.\)

Ta có: 2x + 3y + 4z = 9

=> 2(2k - 1) + 3(4k - 3) + 4(6k - 5) = 9

=> 4k - 2 + 12k - 9 + 24k - 20 = 9

=> (4k + 12k + 24k) - 31 = 9

=> 40k - 31 = 9

=> 40k = 9 + 31

=> 40k = 40

=> k = 1

*Với k = 1 ta có:

x = 2.1 - 1 = 1

y = 4.1 - 3 = 1

z = 6.1 - 5 = 1

Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\)

=> \(\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)

Ta có: x² - y² = 1

=> (5k)² - (4k)² = 1

=> 25k² - 16k² = 1

=> (25 - 16).k² = 1

=> 9k² = 1

=> k² = \(\dfrac{1}{9}\) => \(\left[{}\begin{matrix}k=\dfrac{1}{3}\\k=\dfrac{-1}{3}\end{matrix}\right.\)

*Với k = \(\dfrac{1}{3}\) ta có:

x = \(\dfrac{1}{3}\).5 = \(\dfrac{5}{3}\)

y = \(\dfrac{1}{3}\).4 = \(\dfrac{4}{3}\)

*Với k = \(\dfrac{-1}{3}\) ta có:

x = \(\dfrac{-1}{3}\).5 = \(\dfrac{-5}{3}\)

y = \(\dfrac{-1}{3}\).4 = \(\dfrac{-4}{3}\)

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}\) và \(5x^2+y^2-z^2=117\)

Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}=k\)

=> \(\left\{{}\begin{matrix}x=3k\\y=2k\\z=6k\end{matrix}\right.\)

Ta có:

5x² + y² - z² = 117

=> 5(3k)² + (2k)² - (6k)² = 117

=> 5.9k² + 4k² - 36k² = 117

=> 45k² + 4k² - 36k² = 117

=> (45 + 4 - 36).k² = 117

=> 13k² = 117

=> k² = 117 : 13

=> k² = 9 => \(\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

*Với k = 3 ta có:

x = 3.3 = 9 ; y = 3.2 = 6 ; z = 3.6 = 18

*Với k = -3 ta có:

x = -3.3 = -9 ; y = -3.2 = -6 ; z = -3.6 = -18

a) |2,5 - x| = 1,3 |2,5 - x| = 1,3

=> |2,5 - x| = |2,5 - x| = 1,3 : 1,3

=> |2,5 - x| = |2,5 - x| = 1

=> \(\left[{}\begin{matrix}2,5-x=1\\2,5-x=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1,5\\x=3,5\end{matrix}\right.\)

Vậy x = 1,5 hoặc x = 3,5

b) 1,6 - |x - 0,2| = 0

=> |x - 0,2| = 1,6

=> \(\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)

Vậy x = 1,8 hoặc x = -1,4

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

=> \(x-\dfrac{1}{2}=0\)

=> \(x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

d) (x - 2)² = 1

=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy x = 3 hoặc x = 1

e) (2x - 1)³ = -8

=> (2x - 1)³ = (-2)³

=> 2x - 1 = -2

=> 2x = -1

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

Vậy \(x=\dfrac{-1}{2}\)

f) \(\left(x+\dfrac{1}{2}\right)^2\) = \(\dfrac{1}{16}\)

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

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

Vậy x = \(\dfrac{-1}{4}\) và x = \(\dfrac{-3}{4}\)