Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

26:

A=12x^2+10x-6x-5-(12x^2-8x+3x-2)

=12x^2+4x-5-12x^2+5x+2

=9x-3

Khi x=-2 thì A=-18-3=-21

25:

b: \(\left(y-3\right)\left(y^2+y+1\right)-y\left(y^2-2\right)\)

=y^3+y^2+y-3y^2-3y-3-y^3+2y

=-2y^2-3

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

Bài 1:

\(A=-x^2-2x+9\)

\(A=-\left(x^2+2x-9\right)\)

\(A=-\left(x^2+2x+1-10\right)\)

\(A=-\left(x+1\right)^2+10\)

Vì \(-\left(x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x+1\right)^2+10\le10\)

\(\Rightarrow Amax=10\Leftrightarrow x=-1\)

\(B=-9x^2+6x+25\)

\(B=-\left(9x^2-6x-25\right)\)

\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)

\(B=-\left(3x-1\right)^2+26\)

Vì \(-\left(3x-1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(3x-1\right)^2+26\le26\)

\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(C=-x^2+x+1\)

\(C=-\left(x^2-x-1\right)\)

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

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

Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(D=-2x^2+3x+1\)

\(D=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)

\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)

\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)

Vì \(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x

\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)

\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)

\(E=-25x^2-10x+7\)

\(E=-\left(25x^2+10x-7\right)\)

\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)

\(E=-\left(5x+1\right)^2+8\)

Vì \(-\left(5x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(5x+1\right)^2+8\le8\)

\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

Bài 2:

\(A=9x^2+6x+4\)

\(A=\left(3x\right)^2+2.3x+1+3\)

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

Vì \(\left(3x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(3x+1\right)^2+3\ge3\)

\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)

\(B=4x^2+4x+12\)

\(B=\left(2x\right)^2+2.2x+1+11\)

\(B=\left(2x+1\right)^2+11\)

Vì \(\left(2x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x+1\right)^2+11\ge11\)

\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)

\(C=x^2+x+3\)

\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

\(D=2x^2+3x+1\)

\(D=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)

\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)

\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)

Vì \(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)

\(E=64x^2+16x+3\)

\(E=\left(8x\right)^2+2.8x+1+2\)

\(E=\left(8x+1\right)^2+2\)

Vì \(\left(8x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(8x+1\right)^2+2\ge2\)

\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

Bài 1:

a: A=x^2-6x+10

=x^2-6x+9+1

=(x-3)^2+1>=1

Dấu = xảy ra khi x=3

b: \(B=3x^2-12x+1\)

=3(x^2-4x+1/3)

=3(x^2-4x+4-11/3)

=3(x-2)^2-11>=-11

Dấu = xảy ra khi x=2

bn ko bik lm hay sao, hay là bn chỉ đăng đề lên thôi

sao nhìu... z p , đăq từq câu 1 thôy nha p

Ôi trời sao lắm thế ít thôi bạn nên tách ra mà bạn cần gấp lắm à

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