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mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{2}{x^2-1}-\frac{x}{x-1}+\frac{1}{x+1}\right)\) Đkxđ : x khác 1 ; x khác -1
\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}:\frac{2-x\left(x+1\right)+x-1}{x^2-1}\)
\(A=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}.\frac{x^2-1}{2-x^2-1+x-1}\)
\(A=\frac{4x}{-x^2+x}=\frac{4x}{x\left(1-x\right)}\)
\(A=\frac{4}{1-x}\)
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)
\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)
\(A=\frac{x^2+2004x+2003}{x^2-x}\)
\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)
\(A=\)\(\left[\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right]\left[\frac{a+1}{\sqrt{a}}\right]\)
\(A=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\) \(\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}\)
\(A=\frac{4a\left(a+1\right)}{a-1}\)
ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)
\(a=2\left(16-15\right)\)
\(a=2\)
khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)
vậy.....
Đặt \(a+b=x\) , \(ab=y\)
Ta có biểu thức cần rút gọn :
\(\frac{1}{x^3}.\frac{x\left(x^2-3y\right)}{y^3}+\frac{3}{x^4}.\frac{x^2-2y}{y^2}+\frac{6}{x^5}.\frac{x}{y}=\frac{x^4-3x^2y+3yx^2-6y^2+6y^2}{x^4y^3}=\frac{x^4}{x^4y^3}=\frac{1}{y^3}=\frac{1}{a^3b^3}\)
\(P=\left(a+1\right)\left(a^2+1\right)\left(a^4+1\right)...\left(a^{32}+1\right)\left(a^{64}+1\right)\)
\(\Leftrightarrow10P=\left(a-1\right)\left(a+1\right)\left(a^2+1\right)...\left(a^{64}+1\right)\)
\(\Leftrightarrow10P=\left(a^2-1\right)\left(a^2+1\right)\left(a^4+1\right)...\left(a^{64}+1\right)\)
\(\Leftrightarrow10P=\left(a^4-1\right)\left(a^4+1\right)...\left(a^{64}+1\right)\)
Tiếp tục rút gọn, ta được : \(10P=a^{128}-1\Leftrightarrow P=\frac{a^{128}-1}{10}=\frac{11^{128}-1}{10}\)