Đặt \(-6x^2+7x+2=0\)

\(\Delta=49+48=97\)

\(x_1=\frac{-7-\sqrt{97}}{-12};x_2=\frac{-7+\sqrt{97}}{-12}\)

Câu 4a:

Ta có: 

\(A=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(A=1\cdot\left(a^2-ab+b^2\right)\)

\(A=a^2-ab+b^2\)

\(A=\left(a^2+2ab+b^2\right)-3ab\)

Mà \(ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}\Rightarrow-3ab\ge-\frac{3}{4}\)

\(\Rightarrow A\ge\left(a+b\right)^2-\frac{3}{4}\)

\(=1^2-\frac{3}{4}=\frac{1}{4}\)

Dấu "=" xảy ra khi: \(a=b=\frac{1}{2}\)

Câu 4b:

Ta có: \(\frac{1}{a^2+1}=\frac{\left(a^2+1\right)-a^2}{a^2+1}==1-\frac{a^2}{a^2+1}\)

\(\ge1-\frac{a^2}{2\sqrt{a^2\cdot1}}=1-\frac{a^2}{2a}=1-\frac{a}{2}\left(Cauchy\right)\)

Tương tự CM được: \(\frac{1}{b^2+1}\ge1-\frac{b}{2}\) ; \(\frac{1}{c^2+1}\ge1-\frac{c}{2}\)

Cộng vế 3 BĐT trên lại ta được:

\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge3-\frac{a+b+c}{2}=3-\frac{3}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

Sử dụng bất đẳng thức AM-GM, ta có 

\(\Sigma_{cyc}\sqrt{a+b^2}=\Sigma_{cyc}\frac{a+b^2}{\sqrt{a+b^2}}=\Sigma_{cyc}\frac{\left(a+b\right)\left(a+b^2\right)}{\left(a+b\right)\sqrt{a+b^2}}\ge\Sigma_{cyc}\frac{2\left(a+b\right)\left(a+b^2\right)}{\left(a+b\right)^2+a+b^2}\)

\(=\Sigma_{cyc}\frac{2\left(a+b\right)\left(a\left(a+b+c\right)+b^2\right)}{\left(a+b\right)^2+a\left(a+b+c\right)+b^2}=\Sigma_{cyc}\frac{2\left(a+b\right)\left(a^2+b^2+ab+ac\right)}{2a^2+2b^2+3ab+ac}\)

Như thế ta chỉ cần chứng minh

                     \(\Sigma_{cyc}\frac{\left(a+b\right)\left(a^2+b^2+ab+ac\right)}{2a^2+2b^2+3ab+ac}\ge a+b+c\)

\(\Leftrightarrow\Sigma_{cyc}a^5b^2+\Sigma_{cyc}a^4b^2c+2\Sigma_{cyc}a^5bc\ge2\Sigma_{cyc}a^3b^3c+2\Sigma_{cyc}a^3b^3c^2\)

\(\Leftrightarrow\Sigma_{cyc}\left(\frac{19}{6}a^5b^2+\frac{4}{19}b^5c^2+\frac{6}{19}c^5a^2-a^3b^2c^2\right)+abc\left(\Sigma_{cyc}a^3b-\Sigma_{cyc}a^2bc\right)+2abc\)\(\left(\Sigma_{cyc}a^4-\Sigma_{cyc}a^2b^2\right)\ge0\)

Bất đẳng thức cuối cùng hiển nhiên đúng nên ta có đpcm.Đẳng thức xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)Hoặc \(a=1,b=c=0\) Và các hoán vị

\(4x^3+2x^2+x=7\Leftrightarrow4x^3+2x^2+x-7=0\)

\(\Leftrightarrow4x^3-4x^2+6x^2-6x+7x-7=0\) (  chỗ này xét nghiệm sao cho BT =0 nhé !) 

\(\Leftrightarrow\left(x-1\right)\left(4x^2+6x+7\right)=0\Leftrightarrow x=1\)

\(4x^3+2x^2+x=7\)

\(\Leftrightarrow4x^3+2x^2+x-7=0\)

\(\Leftrightarrow4x^3-4x^2+6x^2-6x+7x-7=0\)

\(\Leftrightarrow\left(4x^3-4x^2\right)+\left(6x^2-6x\right)+\left(7x-7\right)=0\)

\(\Leftrightarrow4x^2.\left(x-1\right)+6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x^2+6x+7\right)=0\)(1)

Ta có: \(4x^2+6x+7=\left(2x\right)^2+2.\frac{3}{2}.2x+\frac{9}{4}+\frac{19}{4}\)

\(=\left(2x+\frac{3}{2}\right)^2+\frac{19}{4}\)

Vì \(\left(2x+\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(2x+\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

\(\Rightarrow4x^2+6x+7>0\)(2)

Từ (1) và (2) \(\Rightarrow x-1=0\)\(\Leftrightarrow x=1\)

Vậy \(x=1\)

\(\left(x+3\right)^2-25=\left(x+3\right)^2-5^2\)

\(=\left(x+3-5\right)\left(x+3+5\right)=\left(x-2\right)\left(x+8\right)\)

\(\left(x+3\right)^2-25\)

\(=\left(x+3-5\right)\left(x+3+5\right)\)

\(=\left(x-2\right)\left(x+8\right)\)

a) 3x²y + x³ - x + 3xy² + y³ - y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= ( x + y )³ - ( x + y )

= ( x + y )[ ( x + y )² - 1 ]

= ( x + y )( x + y - 1 )( x + y + 1 )

b) 2 - 6x² + 7x < xem lại đề >