All categories

1 year ago

X-1/x-2+x+3/x-4=2/(x-2).(4-x)

Mathematics

1 answer:

Varvara68 [4.7K]1 year ago

3 0

The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

According to the statement

we have given that the equation and we have to prove that the given answer is a correct answer for those equivalent equation.

So, The given expression are:

$\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}$

And we have to prove the answer.

So, For this

$\frac{x-1}{x-2} +\frac{x+3}{x-4}$

$\frac{({x-1}) ({x-4}) +({x+3})({x-2})} {(x-2) (x-4)}$

Then the equation become

$\frac{x^{2} -4x -x +4 + x^{2} -2x + 3x -6 }{(x-2) (x-4)}$

Now solve it then

$2x^{2} - 4x -2 / (x-2) (x-4)$

Now take 2 common from answer then equation become

$\frac{x-1}{x-2} +\frac{x+3}{x-4} = \frac{2}{(x-2).(4-x)}$

Hence proved.

So, The given equation x-1/x-2+x+3/x-4=2/(x-2).(4-x) is correct. the answer is proved.

Learn more about equations here

brainly.com/question/2972832

#SPJ1

You might be interested in

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

A brand new filled can of chicken broth is 10 cm tall and has a radius of 10 cm. Mishka uses some of the broth to cook and now t

Maurinko [17]

we have chicken broth in cylindrical can.

we have given the height radius of cylindrical can.

h=10cm r=10 cm .

and she have 6cm broth height in the tin that means she had used 4 cm of height of broth .

now we need to find volume of two cylinder with h=10 cm and 6cm

and we need to find the difference between the volumes .

volume1= $\pi *r*h_{1} =\pi *(10)^2*10=3.14*1000=3140 cm^3$

volume 2= $\pi *(10)^2*6=1884cm^3$

volume used =volume 1-volume2=3140-1884=1256cm^3 is used

8 0

3 years ago

Read 2 more answers

Find the zeros (roots, x-intercepts) for each po<br> 3. f(x) = 5x^3- 20x

Aliun [14]

Answer:

v211

Step-by-step explanation:

6 0

3 years ago

In ΔLMN, n = 910 inches, ∠L=64° and ∠M=31°. Find the length of l, to the nearest 10th of an inch

mr_godi [17]

Answer:

821

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Write the equation of the line passing through the point (2,-1) with a slope of -3

Alexxandr [17]

Y= 2x-3 i think, but i may be wrong

4 0

3 years ago

Read 2 more answers