1.Tìm x:
96-3(x.1)=42
2.Cho a-b=8.Tính:
a.7a-7b
b.9a-3b+9b-3a
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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(9a^3+3a^2+c\right)\left(\frac{1}{9a}+\frac{1}{3}+c\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow A\le\text{∑}\frac{a\left(\frac{1}{9a}+\frac{1}{3}+c\right)}{\left(a+b+c\right)^2}=\text{∑}\left(\frac{1}{9}+\frac{a}{3}+ac\right)\)
\(=\frac{1}{3}+\frac{a+b+c}{3}+\text{∑}ab\le\frac{1}{3}+\frac{1}{3}+\frac{\left(a+b+c\right)^2}{3}=1\)
Dấu "=" khi \(a=b=c=\frac{1}{3}\)
Áp dụng BĐT AM-GM ta có:
\(9a^3+\frac{1}{3}+\frac{1}{3}\ge3\sqrt[3]{9a^3\cdot\frac{1}{3}\cdot\frac{1}{3}}=3a\)
\(3b^2+\frac{1}{3}\ge2\sqrt{3b^2\cdot\frac{1}{3}}=2b\)
Do đó: \(A\le\text{∑}\frac{a}{3a+2b+c-1}=\frac{a}{2a+b}\left(a+b+c=1\right)\)
\(2A\le\text{∑}\frac{2a}{2a+b}=3-\text{∑}\frac{b}{2a+b}=3-\text{∑}\frac{b^2}{2ab+b^2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(2A\le3-\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)
\(=3-\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\Leftrightarrow A\le1\)
Dấu "=" khi \(a=b=c=\frac{1}{3}\)
a,\(\dfrac{9a^2-16b^2}{4b-3a}=\dfrac{\left(3a-4b\right)\left(3a+4b\right)}{\text{4b-3a}}=-3a-4b\)
b,\(\dfrac{25a^2-30ab+9b^2}{3b-5a}=\dfrac{\left(5a-3b\right)^2}{3b-5a}=3b-5a\)
c,\(\dfrac{27a^3-27a^2+9a-1}{9a^2-6a+1}=\dfrac{27a^3-9a^2-18a^2+6a+3a-1}{9a^2-6a+1}=\dfrac{\left(3a-1\right)\left(9a^2-6a+1\right)}{9a^2-6a+1}=3a-1\)
\(\dfrac{3a+7b}{7a+3b}=1\Leftrightarrow3a+7b=7a+3b\)
\(\Leftrightarrow3a=7a+3b-7b\)
\(\Leftrightarrow3a=7a-4b\)
\(\Leftrightarrow4b=7a-3a\)
\(\Leftrightarrow4b=4a\Leftrightarrow a=b\)
Như vậy \(C=-a+b-1=-a+a-1=0-1=-1\)
1. \(96-3\left(x.1\right)=42\)
\(96-3x=42\)
\(3x=96-42\)
\(3x=54\)
\(x=54:3\)
\(x=18\)
2.Theo bài ra ta có: \(a-b=8\)
a. Ta có: \(7a-7b=7\left(a-b\right)\)
\(=7.8=56\)
Bài 1:
96-3(x.1)=42
3.(x.1)= 96 - 42
3. (x.1)= 54
x.1= 54:3
x.1= 18
x=18:1=18
Vậy x = 18
Bài 2:
a) 7a-7b
Theo đề bài cho a-b=8
⇒ 7.a-7.b = 7. (a-b)=7.8=56