\(3\left(x-2\right)+4=5x-2\left(x-1\right)\\ \Leftrightarrow3x-6+4=5x-2x+2\\ \Leftrightarrow0x=4\left(vôlý\right)\)

Vậy pt vô nghiệm

\(2\left(x-2\right)-3\left(1-2x\right)=5\\ \Leftrightarrow2x-4-3+6x=5\\ \Leftrightarrow8x=12\\ \Leftrightarrow x=\dfrac{3}{2}\)

gợi ý nhé

nhận thấy 2x²+11x+19=2x²+5x+7+6(x+2)

đặt ẩn phụ: căn(2x²+5x+7) = a và 3(x+2)=b

=) pt căn(a²+2b)+a=b (=) b(b-2a-2)=0 rồi giải từng trường hợp

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

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)

ĐKXĐ :  \(-4\le x\le4\)

TA CÓ : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)

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

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

\(\Leftrightarrow x\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)

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

\(\Leftrightarrow x=0\)HOẶC  \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)

VỚI \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)

\(\Leftrightarrow\sqrt{4-x}-2=2\sqrt{x+4}\)

\(\Leftrightarrow4-x+4-4\sqrt{4-x}=4x+16\)

\(\Leftrightarrow8-x-4x-16=4\sqrt{4-x}\)

\(\Leftrightarrow-5x-8=4\sqrt{4-x}\)ĐK : \(-4\le x\le\frac{-8}{5}\)

\(\Leftrightarrow\left[-\left(5x+8\right)\right]^2=16\left(4-x\right)\)

\(\Leftrightarrow25x^2+64+80x=64-16x\)

\(\Leftrightarrow25x^2+96x=0\Leftrightarrow x\left(25x+96\right)=0\)

\(\Leftrightarrow x=0\)HOẶC \(x=\frac{-96}{25}\)(THỎA MÃN ĐK )                                                                               

                                                                                               VẬY PT CÓ 2 NGHIỆM \(x\in\left[0;\frac{-96}{25}\right]\)

P/S : CÁCH CỦA MÌNH KHÁ DÀI VÀ CHI TIẾT QUÁ . BẠN CÓ THỂ THAM KHẢO CÁCH KHÁC NHANH HƠN :>

\(\left(5x-\frac{2}{3}\right)-\frac{2x^2-x}{2}\ge\frac{x\left(1-3x\right)}{3}-\frac{5x}{4}\)

<=> \(\frac{60x-8-6\left(2x^2-x\right)}{12}\ge\frac{4x\left(1-3x\right)-15x}{12}\)

<=> \(60x-8-12x^2+6x\ge4x-12x^2-15x\)

<=> \(47x\ge8\)

<=> \(x\ge\frac{8}{47}\)

a) Ta có: \(\left(x+1\right)^4+\left(x-3\right)^4=0\)

Nhận thấy: \(\hept{\begin{cases}\left(x+1\right)^4\ge0\left(\forall x\right)\\\left(x-3\right)^4\ge0\left(\forall x\right)\end{cases}\Rightarrow}\left(x+1\right)^4+\left(x-3\right)^4\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\) (mâu thuẫn)

=> pt vô nghiệm

b) \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^3-8x^2\right)+\left(4x^2-8x\right)+\left(3x-6\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left[\left(x^3+3x^2\right)+\left(x^2+3x\right)+\left(x+3\right)\right]=0\)

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

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)

=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a,\(\left(x+1\right)^4+\left(x-3\right)^4=0\)

\(x^4-1+x^4-81=0\)

\(2x^4-82=0\)

\(2x^4=82\)

\(x^4=41\)

\(x=\sqrt[4]{41}\)

\(\Rightarrow\)vô nghiệm

Lời giải:
ĐK: \(x\geq \frac{-4}{3}\)

BPT \(\Leftrightarrow x^2+6x+13-2\sqrt{3x+4}-3\sqrt{5x+9}\leq 0\)

\(\Leftrightarrow x^2+x+2(x+2-\sqrt{3x+4})+3(x+3-\sqrt{5x+9})\leq 0\)

\(\Leftrightarrow x(x+1)+2.\frac{(x+2)^2-(3x+4)}{x+2+\sqrt{3x+4}}+3.\frac{(x+3)^2-(5x+9)}{x+3+\sqrt{5x+9}}\leq 0\)

\(\Leftrightarrow x(x+1)+\frac{2x(x+1)}{x+2+\sqrt{3x+4}}+\frac{3x(x+1)}{x+3+\sqrt{5x+9}}\leq 0\)

\(\Leftrightarrow x(x+1)\left[1+\frac{2}{x+2+\sqrt{3x+4}}+\frac{3}{x+3+\sqrt{5x+9}}\right]\leq 0\)

\(\Leftrightarrow x(x+1)\leq 0\)

\(\Leftrightarrow -1\leq x\leq 0\)

Kết hợp với ĐKXĐ suy ra nghiệm của BPT là tất cả các số thực thuộc đoạn \([-1;0]\)

\(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

\(\Leftrightarrow\sqrt{2x-5-6\sqrt{2x-5}+9}+\sqrt{2x-5+2\sqrt{2x-5}+1}=4\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}-3+\sqrt{2x-5}+1=4\\\sqrt{2x-5}-3+\sqrt{2x-5}+1=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}-2=4\\2\sqrt{2x-5}-2=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}=6\\2\sqrt{2x-5}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}=3\\\sqrt{2x-5}=-1\left(L\right)\end{cases}}\)

\(\Leftrightarrow2x-5=9\)

\(\Leftrightarrow x=7\)