Đệ biết là có người làm câu c,d nên xin xí câu e :3

ĐK: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)

\(PT\Leftrightarrow5+\sqrt{x+1}=7\left(x-2\right)\)

\(\Leftrightarrow\sqrt{x+1}=7x-19\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\x+1=49x^2-266x+361\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\49x^2-267x+360=0\end{matrix}\right.\)

\(\Rightarrow x=3\left(tm\right)\)

a/ \(\Leftrightarrow\left\{{}\begin{matrix}9-2x\ge0\\x^2-4x-12=\left(9-2x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{9}{2}\\3x^2-32x+93=0\end{matrix}\right.\)

Phương trình vô nghiệm

b/ \(\Leftrightarrow\left(x+1\right)\sqrt[3]{15x^2-x-1}-\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\sqrt[3]{15x^2-x-1}-x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\\sqrt[3]{15x^2-x-1}-x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt[3]{15x^2-x-1}=x-1\)

\(\Leftrightarrow15x^2-x-1=x^3-3x^2+3x-1\)

\(\Leftrightarrow x^3-18x^2+4x=0\)

\(\Leftrightarrow x\left(x^2-18x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=9\pm\sqrt{77}\\\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ge\frac{1}{2}\)

\(\Leftrightarrow2\left(x-1\right)\sqrt{2x-1}-6\left(x-1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(\sqrt{2x-1}-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{2x-1}-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\2x-1=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

d/ ĐKXĐ: \(1\le x< 3\)

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

\(\Leftrightarrow\sqrt{-x^2+4x-3}=2x-5\) (\(x\ge\frac{5}{2}\))

\(\Leftrightarrow-x^2+4x-3=\left(2x-5\right)^2\)

\(\Leftrightarrow5x^2-24x+28=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2< \frac{5}{2}\left(l\right)\\x=\frac{14}{5}\end{matrix}\right.\)

e/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow5+\sqrt{x+1}=7x-14\)

\(\Leftrightarrow\sqrt{x+1}=7x-19\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\x+1=\left(7x-19\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\49x^2-267x+360=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=\frac{120}{49}< \frac{19}{7}\left(l\right)\end{matrix}\right.\)

5) \(ĐK:x\ge-\frac{3}{2}\)

\(x^3+4x-\left(2x+7\right)\sqrt{2x+3}=0\)

\(\Leftrightarrow\frac{x^3+4x}{2x+7}=\sqrt{2x+3}\Leftrightarrow\frac{x^3+4x}{2x+7}-3=\sqrt{2x+3}-3\)

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

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

(không có nghiệm thực)

Vậy phương trình có 1 nghiệm duy nhất là 3

1) \(Pt\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)( đk: \(x\le-3,x\ge0\)

Đặt \(t=\sqrt{x^2+3x},t\ge0\)

Pt trở thành: \(-t^2-3t+10=0\Leftrightarrow t=2\left(dot\ge0\right)\)

giải \(\sqrt{x^2+3x}=2\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

3) \(x^2+\sqrt{2x^2+4x+3}=6-2x\Leftrightarrow-\sqrt{2x^2+4x+3}=x^2+2x-6\)\(\Leftrightarrow\left(2x^2+4x+3\right)-15=-2\sqrt{2x^2+4x+3}\)

Đặt \(\sqrt{2x^2+4x+3}=t\)(t > 0) thì phương trình trở thành \(t^2-15=-2t\Leftrightarrow t^2+2t-15=0\Leftrightarrow\left(t+5\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=-5\left(L\right)\\t=3\left(tm\right)\end{cases}}\)

Với t = 3 thì \(\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x+3=9\Leftrightarrow2x^2+4x-6=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)Vậy phương trình có tập nghiệm S = {1; -3}

câu 1 ta dùng liên hợp nha bạn

điều kiện \(x\ge-1\)

\(\sqrt{x+1}-1+\sqrt[3]{x^2+1}-1=0\\ \Leftrightarrow\frac{x}{\sqrt{x+1}+1}+\frac{x^2}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1}=0\)

suy ra là \(\left[{}\begin{matrix}x=0\left(n\right)\\\frac{1}{\sqrt{x+1}+1}+\frac{x}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1}=0\left(1\right)\end{matrix}\right.\)

theo mình nghĩ (1) vô nghiệm

vậy x=0 là nghiệm pt

Ta dễ dàng chứng minh BĐT sau:

Với \(a;b>1\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

Thật vậy, BĐT tương đương: \(\frac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(a^2+b^2+2\right)\left(1+ab\right)\ge2a^2b^2+2a^2+2b^2+2\)

\(\Leftrightarrow-a^2-b^2+a^3b+ab^3+2ab-2a^2b^2\ge0\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a-b\right)^2\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(a=b\)

Áp dụng vào bài toán:

\(\frac{1}{1+\left(\sqrt{\frac{x^2+15}{2}}\right)^2}+\frac{1}{1+\left(\sqrt{2\left(x^2+3\right)}\right)^2}\ge\frac{2}{1+\sqrt{\left(x^2+3\right)\left(x^2+16\right)}}\)

Dấu "=" xảy ra khi và chỉ khi:

\(\frac{x^2+15}{2}=2\left(x^2+3\right)\Leftrightarrow x^2=1\Rightarrow x=\pm1\)

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)

\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v