Skillfully calculate 157 times 33 / 156

Skillfully calculate 157 times 33 / 156

157×33/156
=（156+1）×33/156
=156×33/156+1×33/156
=33+33/156
=33 and 33 / 156
=33 and 11 / 52

Ab-b & # 178; divided into a + B △ [- (a-b) &# 178; divided into AB + B & # 178;]

The solution of ab-b & # 178; divided into a + B ^ [- (a-b) &# 178; divided into AB + B & # 178;] = a + B / ab-b & # 178; ^ [AB + B & # 178; / - (a-b) &# 178;] = - (a + b) / b (a-b) ^ [b (a + b) / (a-b) &# 178;] = - (a + b) / b (a-b) × [(a-b) &# 178; (b (a + b)] = - (a-b) / B & # 178

If a = 2 of B, find the value of a & # 178; + B & # 178; a & # 178; - AB + B & # 178; of B

A = 2
a=2b
(a²-ab+b²)/(a²+b²)
=(4b²-2b*b+b²)/(4b²+b²)
=3b²/5b²
=0.6

It is known that a + B = 4, a & # 178; + B & # 178; = 11, ask for (a + b) &# 178; (the teacher asked me to talk in the afternoon, I will do it,
How to speak, how to speak, why to do it at each step,
It is known that a + B = 4, a & # 178; + B & # 178; = 11, is to find (a-b) &# 178;, wrong number

(a+b)² = 4² = 16

4【a-b】²-【a+b】²=?

Original form
=【2（a-b）】²-（a+b）²
=【2（a-b）+（a+b）】【2（a-b）-（a+b）】
=（3a-b）（a-3b）

(a+b)²+(a²-b²)+4(a-b)²

(a+b)²+(a²-b²)+4(a-b)²
=a²+2ab+b²+a²-b²+4a²-8ab+4b²
=6a²-6ab+4b²=2(3a²-3ab+2b²)

A & # 178; - B & # 178; = 4, find (a-b) &# 178; (a + b) &# 178;

a²-b²=4
(a+b)(a-b)=4
（a-b）²（a+b）²
=4^2=16

Ab ≤ (a + B / 2) &# 178; what formula is this

0≤(a-b)²=a²-2ab+b²
Then 4AB ≤ A & # 178; + 2Ab + B & # 178; = (a + b) &# 178;
ab≤[(a+b)/2]²

It is known that the quadratic function y = x2-2kx + K2 + K-2 (1) when the value of the real number k is, the image passes through the origin. 2) when the value of the real number k is in what range, the vertex of the function image is in the fourth quadrant

It is known that the quadratic function y = x2-2kx + K2 + K-2 (1) when the value of the real number k is, the image passes through the origin. 2) when the value of the real number k is in what range, the vertex of the function image is in the fourth quadrant

Given quadratic function y = x2-kx + K-5 (1) proof: no matter what real number k takes, the image of the quadratic function and X-axis have two intersections; (2) if the symmetry axis of the image of the quadratic function is x = 1, find its analytical formula

(1) It is proved that: let y = 0, then x2-kx + K-5 = 0, ∵ (?) = k2-4 (K-5) = k2-4k + 20 = (K-2) 2 + 16, ∵ (K-2) 2 ≥ 0, ∵ (K-2) 2 + 16 > 0 ∵ no matter what the real number k is, the image of the quadratic function has two intersections with the x-axis. (2) ∵ the symmetry axis is x = − K2 = K2 = 1, ∵ k = 2, ∵ solution