Bài 3: Nhân, chia số hữu tỉ

NGOC

\(2x.\left(x-\frac{1}{7}\right)=0\)

Ai nhanh cho k nhé

Dương Bá Gia Bảo (2)
18 tháng 7 2019 lúc 10:31

ko viết lại đề

=> \(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0:2=0\\x=0+\frac{1}{7}=\frac{1}{7}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\frac{1}{7}\right\}\)

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Ngọc Lan Tiên Tử
18 tháng 7 2019 lúc 10:32

thôi xong phát hiện ra mình dùng nhầm nick :(

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super UFO.Star
18 tháng 7 2019 lúc 11:04

Ta có: 2x. (x - \(\frac{1}{7}\)) = 0 \(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\) . Vậy x\(\in\left\{0;\frac{1}{7}\right\}\)

hihiChúc bạn học tốt!hehe

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Vũ Minh Tuấn
18 tháng 7 2019 lúc 11:23

\(2x.\left(x-\frac{1}{7}\right)=0\)

=> \(\left\{{}\begin{matrix}2x=0\\x-\frac{1}{7}=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=0:2\\x=0+\frac{1}{7}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)

Vậy \(x\in\left\{0;\frac{1}{7}\right\}\).

Chúc bạn học tốt!

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phamhongson
18 tháng 7 2019 lúc 16:04

x2-2x-17=0 Two solutions were found : x =(2-√72)/2=1-3√ 2 = -3.243 x =(2+√72)/2=1+3√ 2 = 5.243

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2".

Step by step solution : Step 1 : Trying to factor by splitting the middle term

1.1 Factoring x2-2x-17

The first term is, x2 its coefficient is 1 .
The middle term is, -2x its coefficient is -2 .
The last term, "the constant", is -17

Step-1 : Multiply the coefficient of the first term by the constant 1-17 = -17

Step-2 : Find two factors of -17 whose sum equals the coefficient of the middle term, which is -2 .

-17 + 1 = -16
-1 + 17 = 16


Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored

Equation at the end of step 1 : x2 - 2x - 17 = 0 Step 2 : Parabola, Finding the Vertex :

2.1 Find the Vertex of y = x2-2x-17

Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 1.0000

Plugging into the parabola formula 1.0000 for x we can calculate the y -coordinate :
y = 1.0 * 1.00 * 1.00 - 2.0 * 1.00 - 17.0
or y = -18.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-2x-17
Axis of Symmetry (dashed) {x}={ 1.00}
Vertex at {x,y} = { 1.00,-18.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-3.24, 0.00}
Root 2 at {x,y} = { 5.24, 0.00}

Solve Quadratic Equation by Completing The Square

2.2 Solving x2-2x-17 = 0 by Completing The Square .

Add 17 to both side of the equation :
x2-2x = 17

Now the clever bit: Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1

Add 1 to both sides of the equation :
On the right hand side we have :
17 + 1 or, (17/1)+(1/1)
The common denominator of the two fractions is 1 Adding (17/1)+(1/1) gives 18/1
So adding to both sides we finally get :
x2-2x+1 = 18

Adding 1 has completed the left hand side into a perfect square :
x2-2x+1 =
(x-1) (x-1) =
(x-1)2
Things which are equal to the same thing are also equal to one another. Since
x2-2x+1 = 18 and
x2-2x+1 = (x-1)2
then, according to the law of transitivity,
(x-1)2 = 18

We'll refer to this Equation as Eq. #2.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x-1)2 is
(x-1)2/2 =
(x-1)1 =
x-1


Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-1 = 18

Add 1 to both sides to obtain:
x = 1 + √ 18

Since a square root has two values, one positive and the other negative
x2 - 2x - 17 = 0
has two solutions:
x = 1 + √ 18
or
x = 1 - √ 18

Solve Quadratic Equation using the Quadratic Formula

2.3 Solving x2-2x-17 = 0 by the Quadratic Formula .

According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :

- B ± √ B2-4AC
x = ————————
2A


In our case, A = 1
B = -2
C = -17


Accordingly, B2 - 4AC =
4 - (-68) =
72


Applying the quadratic formula :

2 ± √ 72
x = —————
2


Can 72 be simplified ?

Yes! The prime factorization of 72 is
2•2•2•3•3
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

72 = √ 2•2•2•3•3 =2•3•√ 2 =
± 6 • √ 2


2 , rounded to 4 decimal digits, is 1.4142
So now we are looking at:
x = ( 2 ± 6 • 1.414 ) / 2

Two real solutions:

x =(2+√72)/2=1+3√ 2 = 5.243

or:

x =(2-√72)/2=1-3√ 2 = -3.243

Two solutions were found : x =(2-√72)/2=1-3√ 2 = -3.243 x =(2+√72)/2=1+3√ 2 = 5.243


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Subscribe to our mailing list Email Address Terms and Topics Simplifying Radicals Solving Quadratic Equations by Completing The Square Solving Quadratic Equations using the Formula Parabola Finding Vertex and X Intercepts Related Links https://www.khanacademy.org/math/arithmetic/exponents-radicals/radical-radicals/v/simplifying-radicals http://www.regentsprep.org/Regents/math/ALGEBRA/AO1/Lsimplify.htm http://www.purplemath.com/modules/radicals.htm https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/completing-the-square-to-solve-quadratic-equations http://www.purplemath.com/modules/quadform.htm http://www.purplemath.com/modules/grphquad4.htm http://www.regentsprep.org/regents/math/algebra/ac4/lpara.htm
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