a: =4x^2-4x+1+9

=(2x-1)^2+9>=9

Dấu = xảy ra khi x=1/2

b: =2(x^2+3x)

=2(x^2+3x+9/4-9/4)

=2(x+3/2)^2-9/2>=-9/2

Dấu = xảy ra khi x=-3/2

c: =x^2-x+1/4-1/4

=(x-1/2)^2-1/4>=-1/4

Dấu = xảy ra khi x=1/2

Mina giúp Shino đây nè:3(lần lượt nhá)

Ta có:\(4x^2-4x+1=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow2x=1\)

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

1/ f(x) = 4x² - 4x + 1

4x² - 4x + 1 = 0

=> 4x² + 2x + 2x + 1 = 0

=> 2x(2x + 1) + (2x + 1) = 0

=> (2x + 1)(2x + 1) = 0

=> (2x + 1)² = 0

=> 2x + 1 = 0

=> 2x = -1

=> x = -1/2

Vậy nghiệm của đa thức f(x) là x = -1/2

\(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot1+1^2=0\)

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

\(\Leftrightarrow x=-\frac{1}{3}\)

Áp dụng 2 hằng đẳng thức:

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

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

P/s : sửa đề

\(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(A=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(A=\left|2x-1\right|+\left|2x-3\right|\)

\(A=\left|1-2x\right|+\left|2x-3\right|\ge\left|1-2x+2x-3\right|=\left|-2\right|=2\)

Vậy min A = 2 khi và chỉ khi ...........................

Sửa một chút : \(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=\left|2x-1\right|+\left|2x-3\right|\)

\(=\left|2x-1\right|+\left|3-2x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có :

\(A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|=\left|2\right|=2\)

Đẳng thức xảy ra khi \(ab\ge0\)

=> \(\left(2x-1\right)\left(3-2x\right)\ge0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-1\ge0\\3-2x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge1\\-2x\ge-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x\le\frac{3}{2}\end{cases}}\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

2. \(\hept{\begin{cases}2x-1\le0\\3-2x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le1\\-2x\le-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{2}\\x\ge\frac{3}{2}\end{cases}}\)( loại )

=> MinA = 2 <=> \(\frac{1}{2}\le x\le\frac{3}{2}\)

\(A=\left|x+3\right|+\left(y-1\right)^{2018}-4\)

Vì \(\left|x+3\right|\)và \(\left(y-1\right)^{2018}\)\(\ge0\forall x;y\)

\(\Rightarrow A\ge4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy.....

\(C=4-\left|3x-5\right|-\left|5y+8\right|\)

\(C=4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\)

Lí luận như câu a) ta có :

\(C\le4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-8}{5}\end{cases}}\)

Vậy,...........

\(A=\left|x+3\right|+\left(y-1\right)^{2018}-4\)

Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-1\right)^{2018}\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}-4\ge-4\forall x;y\)

\(A=-4\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-1\right)^{2018}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy \(A_{min}=-4\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

\(C=4-\left|3x-5\right|-\left|5y+8\right|\)

Ta có: \(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left|5y+8\right|\ge0\forall y\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-\left|3x-5\right|\le0\forall x\\-\left|5y+8\right|\le0\forall y\end{cases}}\)

\(\Rightarrow4-\left|3x-5\right|-\left|5y+8\right|\le4\forall x;y\)

\(C=4\Leftrightarrow\hept{\begin{cases}-\left|3x-5\right|=0\\-\left|5y+8\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{8}{5}\end{cases}}}\)

Vậy \(C_{max}=4\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{8}{5}\end{cases}}\)

Tham khảo nhé~

Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-1\right)^{2018}\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}\ge0\forall x,y\)

\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}-4\ge-4\forall x,y\)

\(\Rightarrow A\ge-4\)

\(A=-4\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-1\right)^{2018}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy MinA=-4\(\Leftrightarrow\)x=-3: y=1

Ta có: \(C=4-\left|3x-5\right|-\left|5y+8\right|\)

\(=4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\)

Vì\(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left|5y+8\right|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|3x-5\right|+\left|5y+8\right|\ge0\forall x,y\)

\(\Rightarrow-\left(\left|3x-5\right|+\left|5y+8\right|\right)\le0\forall x,y\)

\(\Rightarrow4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\ge4\forall x,y\)

\(\Rightarrow C\ge4\)

\(C=4\Leftrightarrow\hept{\begin{cases}\left|3x-5\right|=0\\\left|5y+8\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-8}{5}\end{cases}}}\)

Vậy MaxC=4\(\Leftrightarrow\)x=\(\frac{5}{3}\): y=\(\frac{-8}{5}\)

1) \(3x^2-4x-7=0\)

\(\Leftrightarrow3x^2+3x-7x-7=0\)

\(\Leftrightarrow3x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\end{cases}}\)

Vậy....

2) \(x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)

Vậy....

3) \(x^3+3x^2+3x+1=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy....

a) ĐKXĐ : \(3\le x\le7\)

Ta có \(A=1.\sqrt{x-3}+1.\sqrt{7-x}\)

\(\le\sqrt{\left(1+1\right)\left(x-3+7-x\right)}=\sqrt{8}\)(BĐT Bunyacovski)

Dấu "=" xảy ra <=> \(\dfrac{1}{\sqrt{x-3}}=\dfrac{1}{\sqrt{7-x}}\Leftrightarrow x=5\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

1, 3x^2 - 4x - 7 =3x^2+3x-7x-7=3x(x+1)-7(x+1)=(3x-7)(x+1)=0

nhiệm là -1 và 7/3

2,x^3-9x=x(x^2-9)=x(x-3)(x+3)=0

nghiệm là 0, 3 và -3

3,x^3+3x^2+3x+1=(x+1)^3=0

nghiệm là -1

Nguyễn Hoàng Long làm kiểu này thì không có được điểm đâu

lm tăt mà thi lại khac

a, x= -0.99996

b, x= -0.286334219

c, x= -0.885584228

a: A=-(x-7)^2-888<=-888

Dấu = xảy ra khi x=7

b: \(B=\left|2x-1\right|+\left|y-5\right|+\dfrac{8}{3}>=\dfrac{8}{3}\)

Dấu = xảy ra khi x=1/2 và y=5

c: \(C=\left(x+3\right)^2+\left|2y-5\right|-232>=-232\)

Dấu = xảy ra khi x=-3 và y=5/2

Bạn cố gắng viết đề bằng công thức toán (công cụ có biểu tượng $\sum$) để được hỗ trợ tốt hơn.