11 + 7 = 14 move a matchstick to make the formula true

11 + 7 = 14 move a matchstick to make the formula true

Place one of 11 across the top of the other and make 11 7

Move a matchstick so that the formula holds 1 + 1 * 1 = 14

Take off the top left one of the four and put it behind the original four
1 + 1 * 1 = 1 + 1

Move only one match
（1） 1×4=44
（2） 1+1×1=14

44 = 44 put 1 under the multiplier
2.1 + 1 × 1 = 1 + 1 move the 4-angled matchstick to the back

Move a match to make the formula true
38×11=306

Move the match on "8" (8 becomes 6) to "0" after the equal sign (0 becomes 9). The result equation is: 36 × 11 = 396,

The first generation of rice weighs 250 kg. They are divided into eight parts on average. How many parts of this bag of rice is each? How many parts of a kilogram? How many parts of a kilogram?

250 △ 8 = 31.25, 125 / 4 kg
31.25 △ 250 = 1 / 8, each is 1 / 8 of this bag of rice,
31.25 △ 1 = 31.25 is 125 / 4 of a kilogram

Find the answer when Lim x tends to infinity (- x) / (2x ^ 2 + 3x-1)

Divide up and down by X & # 178;
The original formula = LIM (- 1 / x) / (2 + 3 / X-1 / X & # 178;)
=0/(2+0-0)
=0

If the value of - 3x & sup2; + MX + NX & sup2; - x + 3 has nothing to do with the value of X, find 2

－3x²+mx+nx²-x+3=(n-3)x²+(m-1)x+3
If x is independent, then (n-3) = 0, n = 3;
(m-1)=0,m=1.
The value of the polynomial is 3

Sixth grade fraction multiplication exercises and application problems
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1. There are 45 students in class 61, including 20 girls. What's the proportion of girls in the class? 2. You can search "YiDianTong teaching website" by Google, and you can find the score multiplication and division problems in unit 3 and 4 of sixth grade mathematics

Find the special solution of y = 0 when y '+ Y / x = SiNx is suitable for x = π

X * dy / DX + y * DX / DX = x * SiNx; D (XY) / DX = x * SiNx; by integrating both sides of X, we can get xy = sinx-x * cosx + C; y = (SiNx) / X - cosx + C / x, where C is any constant. When x = π, y = 0, we can get C = - π, and the special solution is y = (SiNx) / X - cosx - π / X

If the solution set of inequality x2 + (M + 1) x + 1 / 4 M2 > 0 is r, the range of real number m is obtained
Ditto~

Analysis:
If the solution set of inequality X & # 178; + (M + 1) x + 1 / 4 M2 > 0 is r, then:
Δ=(m+1)²-4(1/4)*m²