\(\sqrt{2x^2+3x+2}+\sqrt{4x^2+6x+21}=11\)

Đặt \(\sqrt{2x^2+3x+2}=a;\sqrt{4x^2+6x+21}=b\left(a,b>0\right)\)

Ta có hệ pt :\(\hept{\begin{cases}a+b=11\\b^2-2a^2=17\end{cases}}\)

Đến đây sd pp thế là được nha

Điều kiện: 3x - 2 \(\ge0\) <=> x \(\ge\frac{2}{3}\)

pt <=> \(22x^2-43x^2+43x+x\sqrt{3x-2}-\sqrt{3x-2}-22=0\)

<=> \(\left(22x^2-22\right)+\left(43x-43x^2\right)+\left(x\sqrt{3x-2}-\sqrt{3x-2}\right)=0\)

<=> \(22.\left(x-1\right)\left(x+1\right)+43x\left(1-x\right)+\sqrt{3x-2}.\left(x-1\right)=0\)

<=> \(\left(x-1\right).\left(22x+22-43x+\sqrt{3x-2}\right)=0\)

<=> x-1 = 0 hoặc \(22-21x+\sqrt{3x-2}=0\)

+) x - 1 = 0 => x = 1 (thoả mãn)

+) \(22-21x+\sqrt{3x-2}=0\Leftrightarrow\sqrt{3x-2}=21x-22\) (*)

Điều kiện : 21x - 22 \(\ge\) 0

(*) <=> 3x - 2 = (21x - 22)² <=> 3x - 2 =  441x² - 924x + 484 <=> 441x² - 927x + 486 = 0

Vì 441 - 927 + 486 = 0 => ptrinh có 1 nghiệm x₁ = 1 (loại); x₂ = \(\frac{486}{441}\) (thoả mãn)

vậy phương trình đã cho có 2 nghiệm là: x = 1; x = \(\frac{486}{441}\)

a: =>(x-7)(x+3)=0

hay \(x\in\left\{7;-3\right\}\)

b: =>2x+7=0

hay x=-7/2

c: \(\Delta=50-4\cdot6\cdot2=50-48=2\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{5\sqrt{2}-\sqrt{2}}{12}=\dfrac{\sqrt{2}}{3}\\x_2=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

c.

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)

\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)

\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))

\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)

\(\Rightarrow x^3+7x^2+4x-24=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)

a.

\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)

Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)

Ta có:

\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)

\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)

Đẳng thức xảy ra khi và chỉ khi:

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

Vậy pt có nghiệm duy nhất \(x=3\)

b.

ĐKXD: \(x\ge-1\)

Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)

Đặt \(\sqrt[]{x+1}=t\ge0\)

\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)

\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)

ĐKXĐ: \(x\ge\dfrac{1}{3}\)

PT \(\Leftrightarrow2\left(x-\sqrt{3x-1}\right)+\left[\left(2x+1\right)-\sqrt{3x^2+7x}\right]=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+1\right)}{x+\sqrt{3x-1}}+\dfrac{\left(2x+1\right)^2-\left(3x^2+7x\right)}{2x+1+\sqrt{3x^2+7x}}=0\)

\(\Leftrightarrow\left(x^2-3x+1\right)\left[\dfrac{2}{x+\sqrt{3x-1}}+\dfrac{1}{2x+1+\sqrt{3x^2+7x}}\right]=0\)

Cái ngoặc to vô nghiệm, đến đây bạn có thể giải.