X = 0,5906672909

Hk tốt

\(3x^2+6x-3=\sqrt{\frac{x+7}{3}}\)

\(\Leftrightarrow\left(3x^2+6x-3\right)^2=\left(\sqrt{\frac{x+7}{3}}\right)^2\)

\(\Leftrightarrow9x^4+36x^3+18x^2-36x+9=\frac{x+7}{3}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{69}+7}{6}\\x=\frac{\sqrt{73}-5}{6}\end{cases}}\)

Đặt \(\sqrt{2x^3+7}=a\)

=>6ax=3a^2+1+2x-4a

=>a=2x+1 hoặc a=1/3

=>2x^3+7=(2x+1)^2 hoặc 2x^3+7=1/3

=>\(x\in\left\{1;\dfrac{1-\sqrt{13}}{2};\sqrt[3]{-\dfrac{31}{9}}\right\}\)

ĐK \(0\le x\le1\) và xkhác 1/2 

Đặt \(\sqrt{x}=a;\sqrt{1-x}=b\) ( a>= 0 ; b> = 0 ) => \(a^2+b^2=1\left(1\right)\)

TA có \(6x-3=3\left(2x-1\right)=3\left(a^2-b^2\right)\)

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

Nên pt ban đầu <=> \(\frac{3\left(a^2-b^2\right)}{a-b}=3+2ab\Leftrightarrow3\left(a+b\right)=3+2ab\) (2)

Từ (1) và (2) ta có HPT \(\int^{a^2+b^2=1}_{3\left(a+b\right)=3+2ab}\Leftrightarrow\int^{2ab=\left(a+b\right)^2-1}_{3\left(a+b\right)=3+\left(a+b\right)^2-1\left(I\right)}\)

Đặt a + b = t pt (I) <=> t^2 - 3t + 2 = 0 => t = 1 hoặc 2 

......tự làm tiếp nha 

2x2+5x−1=7x3−1−−−−−√

⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)

Đặt a=x−1−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0

pt (1) trở thành 3a2+2b2−7ab=0

a=2b v a=13b

nha

x=1

                                          Đúng thì  nha 

Đk:\(3\le x\le7\)

Có \(\left(\sqrt{x-3}+\sqrt{7-x}\right)^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4;\forall3\le x\le7\)

\(\Leftrightarrow\sqrt{x-3}+\sqrt{7-x}\ge2\) (I)

Có \(6x-7-x^2=2-\left(x^2-6x+9\right)=2-\left(x-3\right)^2\le2\) (II)

Từ (I) và (II) => Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{\left(x-3\right)\left(7-x\right)}=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow x=3\) (tm)

Vậy...

ĐKXĐ: \(3\le x\le7\)

Ta có:

\(VT=\sqrt{x-3}+\sqrt{7-x}\ge\sqrt{x-3+7-x}=2\)

\(VP=2-\left(x-3\right)^2\le2\)

\(\Rightarrow VT\ge VP\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x-3\right)\left(7-x\right)=0\\\left(x-3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất \(x=3\)

ĐKXĐ: \(3\le x\le7,6x-7-x^2\ge0\)

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

Ta có: \(-x^2+6x-7=-\left(x^2-6x+9\right)+2=-\left(x-3\right)^2+2\le2\)

Ta có: \(\left(\sqrt{x-3}+\sqrt{7-x}\right)^2=x-3+7-x+2\sqrt{\left(x-3\right)\left(7-x\right)}\)

\(=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Rightarrow\sqrt{x-3}+\sqrt{7-x}\ge2\)

\(\Rightarrow\left\{{}\begin{matrix}-x^2+6x-7=2\\\sqrt{x-3}+\sqrt{7-x}=2\end{matrix}\right.\Rightarrow x=3\)

Vậy pt có nghiệm là 3

Dễ thấy có 1 nghiệm x = 0 nên liên hợp đi, trừ 1 ở cả hai vế.

\(\text{ĐK: }\hept{\begin{cases}0\le x\le1\\\sqrt{x}\ne\sqrt{1-x}\end{cases}\Leftrightarrow}\hept{\begin{cases}0\le x\le1\\2x-1\ne0\end{cases}}\)

\(\frac{6x-3}{\sqrt{x}-\sqrt{1-x}}=\frac{3\left(2x-1\right)\left(\sqrt{x}+\sqrt{1-x}\right)}{x-\left(1-x\right)}=\frac{3\left(2x-1\right)\left(\sqrt{x}+\sqrt{1-x}\right)}{2x-1}=3\left(\sqrt{x}+\sqrt{1-x}\right)\)\(\text{Đặt }t=\sqrt{x}+\sqrt{1-x}\)

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

\(\Rightarrow2\sqrt{x-x^2}=t^2-1\)

\(pt\rightarrow3t=3+t^2-1\Leftrightarrow t^2-3t+2=0\Leftrightarrow\orbr{\begin{cases}t=1\\t=2\end{cases}}\)

\(pt\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{cases}}\)