If (M & # 178; - 1) x & # 178; + (M + 1) + 2 = 0 is a linear equation of one variable with respect to x, then M is ()

If (M & # 178; - 1) x & # 178; + (M + 1) + 2 = 0 is a linear equation of one variable with respect to x, then M is ()

Because it is a linear equation of one variable, so M & # 178; - 1 = 0, M + 1 ≠ 0. So m = 1
If the stage is regarded as a straight line, the lighting effect is the best when the lighting device is at two golden section points
Detailed process
The length of the lighting room is known to be 2. 36 meters for stage length
It is known that the length of lighting equipment room is 2.36 meters, and the length of stage is x meters
Both ends of the stage point AB, two golden section points CD = 2.36,
AD=0.618x,AC=AD-CD=0.618x-2.36=DB=AB-AD=x-0.618x
0.618x-2.36=x-0.618x
X = 10 (m)
The second power of x minus x minus 1 = 0, find the value of - 4x + 4x + 9
x²-x-1=0
x²-x=1
-4x²+4x+9
=-4(x²-x)+9
=-4x1+9
=-4+9
=5
The second power of x minus x minus 1 = 0
-4x+4x+9
=-4(x²-x)+9
=-4x1+9
=-4+9
=5
(M & # 178; - 1) x & # 178; - (M + 1) x + 8 = 0. This is a one variable linear equation about X. find M =?
(M & # 178; - 1) x & # 178; - (M + 1) x + 8 = 0, which is the one variable linear equation of X
Ψ M & # 178; - 1 = 0 and M + 1 ≠ 0
Ψ M = ± 1 and m ≠ - 1
∴m=1
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Because it is a linear equation of one variable, M2 - 1 = 0, M + 1 is not equal to 0
So M-1 = 0, M = 1
m²-1 = 0
m = ±1
m+1 ≠ 0
∴ m = 1
Original equation: - (1 + 1) x + 8 = 0
-2x + 8 = 0
x = 4
Given that ab = 10, C and D are two golden section points (AC > BC) on AB, the length of CD is calculated
Golden section of the second day of junior high school
AC=0.618*AB=6.18
BD=0.618*AB=6.18
BC=AD=10-AC=3.82
CD=AB-AD-BC=10-3.82-3.82=2.36
If a + B + c-1-1 = 4a-2 + 2B + 1-4, then the value of a + 2b-3c
Your topic is wrong. It should be a ^ 2 + B ^ 2 + c-1-1 = 4a-2 + 2B + 1-4,
The deformation is: (A-2) ^ 2 ＋ (B-1) ^ 2 ＋ c-1-1 = 0
So: a = 2, B = 1, C = 2
a=2,b=1,c=2
-3c+21=3b+10a
It is known that (M & # 178; - 1) x & # 178; - (M + 1) x + 8 = 0 is a linear equation of one variable with respect to X,
It is known that (M & # 178; - 1) x - (M + 1) x + 8 = 0 is about one variable linear equation, and the value of algebraic formula 199 (M + x) (X-2) + m is obtained
From the meaning of the title: M & # 178; - 1 = 0 and M + 1 is not equal to 0, we get m = 1. Bring in (M & # 178; - 1) x - (M + 1) x + 8 = 0, we get - 2x + 8 = 0, and the solution is x = 4
Substituting M = 1, x = 4 into 199 (M + x) (X-2) + m, the value is 1991
Given that line AB = 4, C is the golden section point of line AB near point B, D is the golden section point of line AC near point C, find the length of CD
zero point six one eight
4*0.618*(1-0.618)=0.944304
1. C from point A: 4 * 0.618 = 2.472
2. D from point A: 2.472 * 0.618 = 1.527696
3. The length of CD point: 2.472-1.527696 = 0.944304
What's the golden section
4*0.618*(1-0.618)=0.944304
1. C from point A: 4 * 0. 618=2。 Four hundred and seventy-two
2. D from point A: 2. 472*0。 618=1。 five hundred and twenty-seven thousand six hundred and ninety-six
3. Length of CD point: 2. 472-1。 527696=0。 nine hundred and forty-four thousand three hundred and four
Given {4A + 8b + 3C = 18 {5A + 10B + 3C = 21, try to find the value of a + 2B + 3C
4a+8b+3c=18 （1）
5a+10b+3c=21 （2）
(2) - (1) get
a＋2b=3
(1) 5 - (2) × 4
3c=6
So a + 2B + 3C = 3 + 6 = 9
It is known that (M & # 178; - 1) x & # 178; - (M + 1 x + 8 = 0) is a linear equation of one variable with respect to X
Find the value of the algebraic formula 199 (M + x) (x-2m) + 3M + 15
The result was 2008
Take a look at the picture and adopt it if you understand,