Câu hỏi của Nguyệt Hà

Question

giải pt$\sqrt{2x^2+x+9}$ +$\sqrt{2x^2-x+1}$ =x+4$\sqrt{\frac{1}{x+3}}$ +$\sqrt{\frac{5}{x+4}}$ =4

tth_new · Accepted Answer

Em có cách này nhưng ko chắc đâu nha!a) ĐK: x>-4Đặt $\sqrt{2x^2+x+9}=a>0;\sqrt{2x^2-x+1}=b>0$ thì:$a^2-b^2=2x+8>0\Rightarrow a>b$ (*)$PT\Leftrightarrow a+b=\frac{a^2-b^2}{2}\Rightarrow2\left(a+b\right)=a^2-b^2$$\Leftrightarrow\left(a-b\right)\left(a+b\right)=2\left(a+b\right)$$\Leftrightarrow\left(a+b\right)\left(a-b-2\right)=0$$\Leftrightarrow\orbr{\begin{cases}a=-b\left(1\right)\a-b=2\left(2\right)\end{cases}}$.*Giải (1): Ta có; a = -b < b (do b >0), mâu thuẫn với (*), loại.*Giải (2): $\Leftrightarrow a=b+2\Leftrightarrow a^2=b^2+4b+4$$\Leftrightarrow2\left(x+4\right)=4\sqrt{2x^2-x+1}+4$$\Leftrightarrow\left(x+2\right)=2\sqrt{2x^2-x+1}$$\Leftrightarrow x^2+4x+4=4\left(2x^2-x+1\right)$$\Leftrightarrow7x^2-8x=0\Leftrightarrow7x\left(x-\frac{8}{7}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\x=\frac{8}{7}\left(TM\right)\end{cases}}$Câu trả lời: Em có cách này nhưng ko chắc đâu nha!a) ĐK: x>-4Đặt $\sqrt{2x^2+x+9}=a>0;\sqrt{2x^2-x+1}=b>0$ thì:$a^2-b^2=2x+8>0\Rightarrow a>b$ (*)$PT\Leftrightarrow a+b=\frac{a^2-b^2}{2}\Rightarrow2\left(a+b\right)=a^2-b^2$$\Leftrightarrow\left(a-b\right)\left(a+b\right)=2\left(a+b\right)$$\Leftrightarrow\left(a+b\right)\left(a-b-2\right)=0$$\Leftrightarrow\orbr{\begin{cases}a=-b\left(1\right)\a-b=2\left(2\right)\end{cases}}$.*Giải (1): Ta có; a = -b < b (do b >0), mâu thuẫn với (*), loại.*Giải (2): $\Leftrightarrow a=b+2\Leftrightarrow a^2=b^2+4b+4$$\Leftrightarrow2\left(x+4\right)=4\sqrt{2x^2-x+1}+4$$\Leftrightarrow\left(x+2\right)=2\sqrt{2x^2-x+1}$$\Leftrightarrow x^2+4x+4=4\left(2x^2-x+1\right)$$\Leftrightarrow7x^2-8x=0\Leftrightarrow7x\left(x-\frac{8}{7}\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\x=\frac{8}{7}\left(TM\right)\end{cases}}$

tth_new · Answer

Note: Em ko chắc nha!b)ĐK: x>-3PT$\Leftrightarrow2-\sqrt{\frac{1}{x+3}}+2-\sqrt{\frac{5}{x+4}}=0$$\Leftrightarrow\frac{4-\frac{1}{x+3}}{2+\sqrt{\frac{1}{x+3}}}+\frac{4-\frac{5}{x+4}}{2+\sqrt{\frac{5}{x+4}}}=0$$\Leftrightarrow\frac{4\left(x+\frac{11}{4}\right)}{\left(x+3\right)\left(2+\sqrt{\frac{1}{x+3}}\right)}+\frac{4\left(x+\frac{11}{4}\right)}{\left(x+4\right)\left(2+\sqrt{\frac{5}{x+4}}\right)}=0$$\Leftrightarrow\left(x+\frac{11}{4}\right)\left[\frac{4}{\left(x+3\right)\left(2+\sqrt{\frac{1}{x+3}}\right)}+\frac{4}{\left(x+4\right)\left(2+\sqrt{\frac{5}{x+4}}\right)}\right]=0$Cái ngoặc to lớn hơn 0 (hiển nhiên)Bí.Câu trả lời: Note: Em ko chắc nha!b)ĐK: x>-3PT$\Leftrightarrow2-\sqrt{\frac{1}{x+3}}+2-\sqrt{\frac{5}{x+4}}=0$$\Leftrightarrow\frac{4-\frac{1}{x+3}}{2+\sqrt{\frac{1}{x+3}}}+\frac{4-\frac{5}{x+4}}{2+\sqrt{\frac{5}{x+4}}}=0$$\Leftrightarrow\frac{4\left(x+\frac{11}{4}\right)}{\left(x+3\right)\left(2+\sqrt{\frac{1}{x+3}}\right)}+\frac{4\left(x+\frac{11}{4}\right)}{\left(x+4\right)\left(2+\sqrt{\frac{5}{x+4}}\right)}=0$$\Leftrightarrow\left(x+\frac{11}{4}\right)\left[\frac{4}{\left(x+3\right)\left(2+\sqrt{\frac{1}{x+3}}\right)}+\frac{4}{\left(x+4\right)\left(2+\sqrt{\frac{5}{x+4}}\right)}\right]=0$Cái ngoặc to lớn hơn 0 (hiển nhiên)Bí.

FL.Hermit · Answer

Câu 2 nhé: (CÁCH NÀY TUI NGHĨ ỔN ĐỊNH HƠN)pt <=> $\frac{1}{x+3}+\frac{5}{x+4}+2\sqrt{\frac{5}{\left(x+3\right)\left(x+4\right)}}=16$<=> $\frac{1}{x+3}+\frac{5}{x+4}-16=-2\sqrt{\frac{5}{\left(x+3\right)\left(x+4\right)}}$<=> $\frac{x+4+5\left(x+3\right)-16\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}=-2\sqrt{\frac{5}{\left(x+3\right)\left(x+4\right)}}$<=> $\frac{\left(x+4+5\left(x+3\right)-16\left(x+3\right)\left(x+4\right)\right)^2}{\left(x+3\right)^2\left(x+4\right)^2}=\frac{20}{\left(x+3\right)\left(x+\text{4}\right)}$<=> $\left(16\left(x+3\right)\left(x+4\right)-\left(x+4\right)-5\left(x+3\right)\right)^2=20\left(x+3\right)\left(x+4\right)$Ta đặt: $x+4=a;x+3=b$ => $a-b=1$ => $a=b+1$=> TA CÓ PT MS: $\left(16ab-a-5b\right)^2=20ab$<=> $\left(16b\left(b+1\right)-b-1-5b\right)^2=20b\left(b+1\right)$<=> $\left(16b^2+10b-1\right)^2=20b^2+20b$<=> $256b^4+100b^2+1+320b^3-32b^2-20b=20b^2+20b$<=> $256b^4+320b^3+48b^2-40b+1=0$Bạn ấn nghiệm ra thì có 1 no $=\frac{1}{4}$; 1 nghiệm vô tỉ và do $b=x+3\left(cmt\right)$=> TA DỄ DÀNG TÌM RA X.Câu trả lời: Câu 2 nhé: (CÁCH NÀY TUI NGHĨ ỔN ĐỊNH HƠN)pt <=> $\frac{1}{x+3}+\frac{5}{x+4}+2\sqrt{\frac{5}{\left(x+3\right)\left(x+4\right)}}=16$<=> $\frac{1}{x+3}+\frac{5}{x+4}-16=-2\sqrt{\frac{5}{\left(x+3\right)\left(x+4\right)}}$<=> $\frac{x+4+5\left(x+3\right)-16\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+4\right)}=-2\sqrt{\frac{5}{\left(x+3\right)\left(x+4\right)}}$<=> $\frac{\left(x+4+5\left(x+3\right)-16\left(x+3\right)\left(x+4\right)\right)^2}{\left(x+3\right)^2\left(x+4\right)^2}=\frac{20}{\left(x+3\right)\left(x+\text{4}\right)}$<=> $\left(16\left(x+3\right)\left(x+4\right)-\left(x+4\right)-5\left(x+3\right)\right)^2=20\left(x+3\right)\left(x+4\right)$Ta đặt: $x+4=a;x+3=b$ => $a-b=1$ => $a=b+1$=> TA CÓ PT MS: $\left(16ab-a-5b\right)^2=20ab$<=> $\left(16b\left(b+1\right)-b-1-5b\right)^2=20b\left(b+1\right)$<=> $\left(16b^2+10b-1\right)^2=20b^2+20b$<=> $256b^4+100b^2+1+320b^3-32b^2-20b=20b^2+20b$<=> $256b^4+320b^3+48b^2-40b+1=0$Bạn ấn nghiệm ra thì có 1 no $=\frac{1}{4}$; 1 nghiệm vô tỉ và do $b=x+3\left(cmt\right)$=> TA DỄ DÀNG TÌM RA X.

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