Đk: \(5x^2+10x+1\ge0\)

Đặt \(t=\sqrt{5x^2+10x+1}\ge0\)

\(pt\Leftrightarrow\sqrt{5x^2+10x+1}=\frac{-\left(5x^2+10x+1\right)}{5}+\frac{36}{5}\)

\(\Leftrightarrow5t=-t^2+36\Leftrightarrow t^2+5t-36=0\)

\(\Leftrightarrow\left(t-4\right)\left(t+9\right)=0\Leftrightarrow t=4\) ( do \(t\ge0\) )

\(\Leftrightarrow5x^2+10x+1=16\Leftrightarrow5x^2+10x-15=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)( TM )

3x²+6x+7=3.(x²+2x+1)+4=3.(x+1)²+4 >= 4 

=> căn của nó >=

..................................................... ko thích giải

dùng BĐT nha bạn

ghê quá, muốn ói , mk chưa học đâu

a)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)

\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)

Vậy pt có một nghiệm duy nhất là \(x=-1\)

b)

\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)

\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)

\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)

Lập bảng xét dấu ra nhé ~^o^~

a/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\sqrt{x+1}=1+\sqrt{x-2}\)

\(\Leftrightarrow x+1=1+x-2+2\sqrt{x-2}\)

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

\(\Leftrightarrow x=3\)

b/ ĐKXĐ: \(x^2\ge2\)

Đặt \(\sqrt{x^2-2}=t\ge0\Rightarrow x^2=t^2+2\)

Pt trở thành: \(t^2+2-t=4\)

\(\Leftrightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-2}=2\Leftrightarrow x^2=6\Rightarrow x=\pm\sqrt{6}\)

c/

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=5-\left(x+1\right)^2\)

Do \(\left(x+1\right)^2\ge0\) ;\(\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{3\left(x+1\right)^2+4}\ge\sqrt{0+4}=2\\\sqrt{5\left(x+1\right)^2+9}\ge\sqrt{0+9}=3\end{matrix}\right.\)

\(\Rightarrow VT\ge5\)

\(VP=5-\left(x+1\right)^2\le5\)

\(\Rightarrow VT\ge VP\)

Dấu "=" xảy ra khi và chỉ khi: \(\left(x+1\right)^2=0\Leftrightarrow x=-1\)

a) \(3\sqrt{x^2+3x}=\left(x+5\right)\left(2-x\right)\)

\(\Leftrightarrow3\sqrt{x^2+3x}=-x^2-3x+10\)

\(\Leftrightarrow\left(x^2+3x\right)+3\sqrt{x^2+3x}-10=0\)

Đặt \(t=\sqrt{x^2+3x}\left(t\ge0\right)\left(1\right)\)

Ta có:

\(\Rightarrow t^2+3t-10=0\)

\(\Rightarrow t_1=2\left(TM\right);t_2=-5\left(KTM\right)\)

thay \(t=2\) vào (1), ta có :

\(\sqrt{x^2+3x}=2\)

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

\(\Rightarrow x_1=1;x_2=-4\)

vậy phương trình có 3 nghiệm x1 = 1, x2 = -4

b) \(\sqrt{5x^2+10x+1}=7-x^2-2x\)

\(\Leftrightarrow\sqrt{5x^2+10x+1}=\left(5x^2+10x+1\right)-6x^2+12x-6\)

\(\Leftrightarrow\sqrt{5x^2+10x+1}=\left(5x^2+10x+1\right)-6\left(x-1\right)^2\)

Đặt \(t=\sqrt{5x^2+10x+1}\) (t lớn hơn hoặc bằng 0) (1)

ta có :...............

mk chỉ bt làm đến đấy thôi, hình như đây là ôn hsg toán 10 à

a)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

\(pt\Leftrightarrow\sqrt{3x^2+6x+3+4}+\sqrt{5x^2+10x+5+9}=-x^2-2x+4\)

\(\Leftrightarrow\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+9}=-x^2-2x+4\)

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-x^2-2x+4\)

Dễ thấy: \(\hept{\begin{cases}3\left(x+1\right)^2\ge0\\5\left(x+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}3\left(x+1\right)^2+4\ge4\\5\left(x+1\right)^2+9\ge9\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{3\left(x+1\right)^2+4}\ge2\\\sqrt{5\left(x+1\right)^2+9}\ge3\end{cases}}\)

\(\Rightarrow VT=\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}\ge2+3=5\)

Và \(VP=-x^2-2x+4=-x^2-2x-1+5\)

\(=-\left(x^2+2x+1\right)+5=-\left(x+1\right)^2+5\le5\)

SUy ra \(VT\ge VP=5\Leftrightarrow x=-1\)

b)\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)

\(pt\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1}=1\)

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

..... giải nốt tiếp ra x=1

c)Sửa đề \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)

ĐK:....

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{x-7}+\sqrt{9-x}\right)^2\)

\(\le\left(1+1\right)\left(x-7+9-x\right)=4\)

\(\Rightarrow VT^2\le4\Rightarrow VT\le2\)

Lại có: \(VP=x^2-16x+66=x^2-16x+64+2\)

\(=\left(x-8\right)^2+2\ge2\)

Suy ra \(VT\ge VP=2\) khi \(VT=VP=2\)

\(\Rightarrow\left(x-8\right)^2+2=2\Rightarrow x-8=0\Rightarrow x=8\)