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Trả lời:
1) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=\left(\sqrt{x}\right)^2-2\sqrt{x}+\sqrt{x}-2=x-\sqrt{x}-2\)
2) \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2=x^2-2x+4x-8-\left(x^2-6x+9\right)\)\(=x^2+2x-8-x^2+6x-9=8x-17\)
3) \(3x\left(2x^3-3x^2+5\right)=6x^4-9x^3+15x\)
\(1,\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)
\(x-2\sqrt{x}-3\sqrt{x}+6\)
\(x-5\sqrt{x}+6\)
\(2,\left(x+2\right)\left(x-3\right)+x\left(x+1\right)\)
\(x^2+2x-3x-6+x^2+x\)
\(2x^2-6\)
b) lấy kết quả rút gọn của câu A ta được
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}< 1.=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}-1< 0\)
\(P=\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2}{\sqrt{x}-1}\)
đề bài cho x>=0 ta suy ra luôn
\(x+2>0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow x< 1\)
vậy x <1 thì P < 1
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\sqrt{x}\left(x+1\right)-\left(x+1\right)}\right).\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{x+1+\sqrt{x}}{x+1}\right):\left(\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(P=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(x+\sqrt{x}+1\right)}{\left(x+1\right)}.\frac{\left(x+1\right)}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
Bài a,b,c,e,g,i thì đặt điều kiện rồi bình phương 2 vế rồi giải, bài j chuyển vế rồi bình phương
Chỉ trình bày lời giải, tự tìm điều kiện nha :v
d) \(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Rightarrow x-1=1\Leftrightarrow x=2\)
f) \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\sqrt{x-4+2.2\sqrt{x-4}+4}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}=2\)
\(\Leftrightarrow\sqrt{x-4}+2=2\)
\(\Leftrightarrow\sqrt{x-4}=0\)
\(\Rightarrow x-4=0\Leftrightarrow x=4\)
Chậc :))) T còn cách khác đây =)))
\(\sqrt{x-1+2\sqrt{x-2}}-\sqrt{x-1-2\sqrt{x-2}}=1\)
\(\Leftrightarrow\left(\sqrt{x-1+2\sqrt{x-1}}\right)^2=\left(1+\sqrt{x-1-2\sqrt{x-2}}\right)^2\)
\(\Leftrightarrow x-1+2\sqrt{x-2}-x=2\sqrt{x-1-2\sqrt{x-2}}+x-2\sqrt{x-2}-x\)
\(\Leftrightarrow2\sqrt{x-2}-1=2\sqrt{x-1-2\sqrt{x-2}}-2\sqrt{x-2}\)
\(\Leftrightarrow4x-4\sqrt{x-2}-7=-8\sqrt{x-2}-8\sqrt{x-2}.\sqrt{x-2\sqrt{x-2}-1}+8x-12\)
\(\Leftrightarrow5-4\sqrt{x-2}-4x=-8\sqrt{x-2}-8\sqrt{x-2}.\sqrt{x-2\sqrt{x-2}-1}\)
\(\Leftrightarrow x=\frac{9}{4}\) (tmyk)
\(1.\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)+1\left(\sqrt{x}-2\right)\)
\(=x-2\sqrt{x}+\sqrt{x}-2\)
\(=x-\sqrt{x}-2\)
\(2.\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2\)
\(=x\left(x-2\right)+4\left(x-2\right)-\left(x^2-6x+9\right)\)
\(=x^2-2x+4x-8-x^2+6x-9\)
\(=8x-17\)