Cho biểu thức
Ạ =\(\frac{5624x17+82x5624+5624}{1016-720:\left(a-6\right)}\) Tính giá trị biểu thức
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a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
ĐK: \(x-9\ne0\Rightarrow x\ne9\)
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(x+\sqrt{x}-6\ne0\Rightarrow x+3\sqrt{x}-2\sqrt{x}-6\ne0\Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\ne0\)
\(\Rightarrow\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(A=\left(\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{1}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\left(\frac{1+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}:\frac{1+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{4\sqrt{x}-12}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-3\right)}\)
2, Với \(x=\frac{25}{16}\)\(\Rightarrow\sqrt{x}=\sqrt{\frac{25}{16}}=\frac{5}{4}\)
\(A=\frac{\frac{5}{4}\left(\frac{5}{4}-2\right)}{4\left(\frac{5}{4}-3\right)}=\frac{5}{4}.\left(-\frac{3}{4}\right):4\left(-\frac{7}{4}\right)=-\frac{15}{16}:-7=\frac{15}{112}\)
\(\orbr{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)\(\orbr{\begin{cases}\orbr{\begin{cases}\sqrt{x}-2< 0\\\sqrt{x}-3>0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}< 2\\\sqrt{x}>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< 4\\x>9\end{cases}}}\\\orbr{\begin{cases}\sqrt{x}-2>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}>2\\\sqrt{x}< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>4\\x< 9\end{cases}}}}\end{cases}}\)
\(\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right)=\frac{a^2-1}{a^2-a}=\frac{a+1}{a}\)
ở phàn a+/a thiếu số 1 nhé
\(\frac{1}{a+1}+\frac{2}{a^2-1}=\frac{a-1+2}{a^2-1}=\frac{1}{a-1}\)
=> K =\(\frac{a^2-1}{a}\)
đkxđ: a khác +-1
b, thay vào mà tình
a/ \(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)\)
\(=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\frac{a^2-1}{a\left(a-1\right)}:\frac{a-1+2}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}.\frac{\left(a-1\right)\left(a+1\right)}{a-1}\)
\(=\frac{a+1}{a}.a+1\)
\(=\frac{\left(a+1\right)^2}{a}\)
b, Thay a=1/2
\(\Rightarrow\frac{\left(\frac{1}{2}+1\right)^2}{\frac{1}{2}}=\frac{\frac{9}{4}}{\frac{1}{2}}=\frac{9}{2}\)
C1: dễ nên tự làm nhé
C2: \(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{2}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=6-5-3+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)\)
\(=-2-\frac{1}{2}=\frac{-4}{2}-\frac{1}{2}=\frac{-5}{2}\)
ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
với ĐKXĐ ta có
=\(\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{7\left(x-1\right)}\)
=\(\frac{4x}{\left(x+1\right)\left(x-1\right)}\times\frac{7\left(x-1\right)}{2x}\)
=\(\frac{14}{x+1}\)
b, x=6(t/m)
khi x=6 thì A=\(\frac{14}{6+1}=2\)
c,A=7<=>\(\frac{14}{x+1}=7\)
\(\Leftrightarrow7x+7=14\)
\(\Leftrightarrow7x=7\Leftrightarrow x=1\left(loại\right)\)
Vậy ko có giá trị x để A=7
A = 5(x + 3)(x - 3) + (2x + 3)3 + (x - 6)2
A = 5(x + 3)(x - 3) + 4x2 + 12x + 9 + x2 - 12x + 36
A = 5x2 - 45x + 4x2 + 12x + 9 + x2 - 12x + 36
A = 10x2 (1)
Thay x = -1/5 vào (1), ta có:
A = 10x2 = 10.(-1/5)2 = 2/5
A = 2/5
Vậy:...