Đặt \(A=3\left(x^2-\dfrac{8}{3}x\right)-2=3\left(x^2-2.\dfrac{4}{3}x+\dfrac{16}{9}-\dfrac{16}{9}\right)-2\)

\(=3\left(x^2-2.\dfrac{4}{3}x+\dfrac{16}{9}\right)-\dfrac{16}{3}-2\)

\(=3\left(x-\dfrac{4}{3}\right)^2-\dfrac{22}{3}\ge-\dfrac{22}{3}\)

\(A_{min}=-\dfrac{22}{3}\) khi \(x=\dfrac{4}{3}\)

Ta có: \(3x^2-8x-2\)

\(=3\left(x^2-\dfrac{8}{3}x-\dfrac{2}{3}\right)\)

\(=3\left(x^2-2\cdot x\cdot\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{22}{9}\right)\)

\(=3\left(x-\dfrac{4}{3}\right)^2-\dfrac{22}{3}\ge-\dfrac{22}{3}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{4}{3}\)

Đặt \(f\left(x\right)=-x^2-2x-3\)

\(=-x^2-x-x-3\)

\(=-x.\left(x-1\right)-\left(x-1\right)-2\)

\(=-[-\left(x-1\right)^2]-2\le-2< 0\)

\(\Rightarrow\)Đa thức không có nghiệm

Đặt \(A=-x^2-2x-3\)

\(\Rightarrow-A=x^2+2x+3\)

\(-A=\left(x^2+2x+1\right)+2\)

\(-A=\left(x+1\right)^2+2\)

\(\Rightarrow A=-\left(x+1\right)^2-2\)

Ta có: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+1\right)^2-2\le2\forall x\)

\(\Rightarrow\) Đa thức vô nghiệm

* Tìm GTNN : 

\(a)\) Đặt \(A=4x^2+4x+2\) ta có : 

\(A=\left(4x^2+4x+1\right)+1\)

\(A=\left[\left(2x\right)^2+2.2x.1+1^2\right]+1\)

\(A=\left(2x+1\right)^2+1\ge1\)

Dấu "=" xay ra \(\Leftrightarrow\)\(\left(2x+1\right)^2=0\)

\(\Leftrightarrow\)\(2x+1=0\)

\(\Leftrightarrow\)\(2x=-1\)

\(\Leftrightarrow\)\(x=\frac{-1}{2}\)

Vậy GTNN của \(A\) là \(1\) khi \(x=\frac{-1}{2}\)

Bài b) là tìm GTLN thì đúng hơn 

\(b)\) Đặt \(B=-2x^2-16x-31\) ( hình như đề thiếu chỗ \(16x\) ) ta có : 

\(-2B=4x^2+32x+62\)

\(-2B=\left(4x^2+32x+64\right)-2\)

\(-2B=\left[\left(2x\right)^2+2.2x.8+64\right]-2\)

\(-2B=\left(2x+8\right)^2-2\ge-2\)

\(B\le\frac{-2}{-2}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(2x+8\right)^2=0\)

\(\Leftrightarrow\)\(2x+8=0\)

\(\Leftrightarrow\)\(2x=-8\)

\(\Leftrightarrow\)\(x=\frac{-8}{2}\)

\(\Leftrightarrow\)\(x=-4\)

Vậy GTLN của \(B\) là \(1\) khi \(x=-4\)

Chúc bạn học tốt ~ 

a) ĐKXĐ : 3 – 4x ≠ 0 và 3 + 4x ≠ 0 (16x2 – 9 = - (3 – 4x)(3 + 4x) ≠ 0)

⇔ x ≠ 3/4 và x ≠ -3/4

Quy đồng mẫu thức :

Khử mẫu, ta được :

-12x2 – 30x + 21 – (9x + 12x2 – 21 – 28x) = 18x – 24x2 + 15 – 20x

⇔ -12x2 – 30x + 21 – 9x – 12x2 + 21 + 28x = 18x – 24x2 + 15 – 20x

⇔ -9x = -27 ⇔ x = 3 (thỏa mãn ĐKXĐ)

Tập nghiệm : S = {3}

b) (x + 3)2 - (x -3)2 = 6x + 18

⇔ x2 + 6x + 9 – x2 + 6x – 9 = 6x + 18

⇔ 6x = 18 ⇔ x = 3

Tập nghiệm : S = {3}

a) Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

a) \(8x^2+30x+7=0\)

\(\Leftrightarrow8\left(x^2+\frac{15}{4}x+7\right)=0\)

\(\Leftrightarrow x^2+\frac{1}{4}x+\frac{7}{2}x+\frac{7}{8}=0\) 

\(\Leftrightarrow x\left(x+\frac{1}{4}\right)+\frac{7}{2}\left(x+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(x+\frac{1}{4}\right)\left(x+\frac{7}{2}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{7}{2}\end{array}\right.\)

b)\(x^3-11x^2+30x=0\)

\(\Leftrightarrow x\left(x^2-11x+30\right)=0\)

\(\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)

\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\\x-6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\\x=6\end{array}\right.\)

a) \(\Delta=\left(30\right)^2-4.8.7=676>0\) ( PTC2NPB )

\(X_1=\frac{-30+\sqrt{676}}{16}\)

\(X_2=\frac{-30-\sqrt{676}}{16}\)

\(30x-15x^2-0\)

\(\Leftrightarrow15x\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}15x=0\\2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

13x² + 9y² - 30x + 12xy + 25 = 0

<=> (9y² + 12xy + 4y²) + (9x² - 30x + 25) = 0

<=> (3y + 2x)² + (3x - 5)² = 0

Dễ thấy \(\left(3y+2x\right)^2\ge0;\left(3x-5\right)^2\ge0\forall x,y\)

nên \(\left(3y+2x\right)^2+\left(3x-5\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3y+2x=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{10}{9}\\x=\dfrac{5}{3}\end{matrix}\right.\)

Bạn xem lại đề

\(9x^2+4y^2+26+4y=30x\)

\(\Leftrightarrow9x^2-30x+4y^2+4y+26=0\)

\(\Leftrightarrow\left(9x^2-30x+25\right)+\left(4y^2+4y+1\right)=0\) 

\(\Leftrightarrow\left(3x-5\right)^2+\left(2y+1\right)^2=0\)

Mà: \(\left\{{}\begin{matrix}\left(3x-5\right)^2\ge0\forall x\\\left(2y+1\right)^2\ge0\forall x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=5\\2y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\left(x^2-x-30\right)=0\)

\(\Leftrightarrow x\left(x-6\right)\left(x+5\right)=0\)

hay \(x\in\left\{0;6;-5\right\}\)