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3 years ago

Solve for x. 3/x-1 - 1/x^-1=5/x-1 Please show all work

Mathematics

2 answers:

navik [9.2K]3 years ago

5 0

3/(x-1) - 1/(x^2-1) = 5/(x-1)

Subtract 3/(x-1) from both sides

-1/(x^2-1) = 2/(x-1)

Factor x^2 - 1

-1/[(x-1)(x+1] = 2/(x-1)

Multiply by (x-1)(x+1) on both sides

-1 = 2 (x+1)

-1 = 2x + 2

Subtract 2 from both sides

-3 = 2x

Divide by 2 on both sides

-3/2 = x

Kitty [74]3 years ago

3 0

Answer:

x = -3/2

Step-by-step explanation:

$\frac{3(x+1)}{(x-1)(x+1)} - \frac{1}{(x-1)(x+1)} = \frac{5(x+1)}{(x-1)(x+1)} \\$ Find the common denominator

3(x+1) - 1 = 5(x+1) Distribute

3x + 3 -1 = 5x +5 Combine Like terms

3x + 2 = 5x + 5 Solve for x

2 = 2x +5

-3 = 2x

x = -3/2

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Step-by-step explanation:

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3 years ago

Here its hard and i need help thank you

vagabundo [1.1K]

Answer:

the volume of a cylinder = V=Ah =pirsqh

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4 0

3 years ago

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Futhe Mathematics <img src="https://tex.z-dn.net/?f=%28Cos%20%7B%7D%5E%7B4%7Dt%20-Sin%20%7B%7D%5E%7B4%7Dt%20%29%20%20%5Cdiv%2

Nastasia [14]

Answer:

cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

$\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }$

We can write the numerator as ,

$\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }$

Recall the identity ,

$\longrightarrow (a-b)(a+b)=a^2-b^2$

Using this we have ,

$\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}$

Again , as we know that ,

$\longrightarrow sin^2\phi + cos^2\phi = 1$

Therefore we can rewrite it as ,

$\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}$

Again using the first identity mentioned above ,

$\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}$

Or else we can also write it using ,

$\longrightarrow cos2\phi = cos^2\phi - sin^2\phi$

Therefore ,

$\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}$

And we are done !

$\rule{200}{4}$

Additional info :-

Derivation of cos²x - sin²x = cos2x :-

We can rewrite cos 2x as ,

$\longrightarrow cos(x + x )$

As we know that ,

$\longrightarrow cos(y + z )= cosy.cosz - siny.sinz$

So that ,

$\longrightarrow cos(x+x) = cos(x).cos(x) - sin(x)sin(x)$

On simplifying,

$\longrightarrow cos(x+x) = cos^2x - sin^2x$

Hence,

$\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}$

$\rule{200}{4}$

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3 years ago

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SIZIF [17.4K]

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