\(P=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)

\(=\frac{\left(x+\frac{1}{x}\right)^6-\left[\left(x^3\right)^2+2x^3\cdot\frac{1}{x^3}+\left(\frac{1}{x^3}\right)^2\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)

\(=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)

\(=\frac{\left[\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\right]\left[\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\right]}{\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)}\)

\(=\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)\ge\left(2\sqrt{x\cdot\frac{1}{x}}\right)^3+2\sqrt{x^3\cdot\frac{1}{x^3}}=8+2=10\)

Dấu "=" khi x = 1

1)

ĐKXĐ: x\(\ne\)3

ta có :

\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)

để biểu thức A có giá trị = 1

thì :\(\frac{x-3}{2}\)=1

=>x-3 =2

=>x=5(thoả mãn điều kiện xác định)

vậy để biểu thức A có giá trị = 1 thì x=5

1)

\(A=\frac{x^2-6x+9}{2x-6}\)

A xác định

\(\Leftrightarrow2x-6\ne0\)

\(\Leftrightarrow2x\ne6\)

\(\Leftrightarrow x\ne3\)

Để A = 1

\(\Leftrightarrow x^2-6x+9=2x-6\)

\(\Leftrightarrow x^2-6x-2x=-6-9\)

\(\Leftrightarrow x^2-8x=-15\)

\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)

\(\frac{-1}{6}.\frac{-15}{19}.\frac{38}{45}\)

\(\frac{\left(-1\right).\left(-15\right).2.19}{2.3.19.3.15}\)

\(=\frac{1}{3.3}=\frac{1}{9}\)

\(\left(-\frac{1}{6}\right).\left(-\frac{15}{19}\right).\frac{38}{45}\)

\(=\frac{\left(-1\right).\left(-15\right).38}{6.19.45}=\frac{1.15.38}{6.19.45}\)

\(=\frac{1.1.1}{3.1.3}=\frac{1}{9}\)

\(\frac{-1.\left(-15\right).38}{6.19.45}\)=\(\frac{-1.\left(-3\right).2}{6.9}\)=\(\frac{-1\left(-1\right).2}{2.9}\)=\(\frac{1}{9}\)

a,\(\frac{4}{9}.\frac{2}{6}=\frac{4}{27}\)

b,\(1\frac{1}{3}.\left(0,5\right)+\left(\frac{8}{15}-\frac{19}{30}\right):\frac{6}{15}\)

=\(\frac{4}{3}.\frac{1}{2}+\left(\frac{16}{30}-\frac{19}{30}\right).\frac{15}{6}\)

=\(\frac{2}{3}+\frac{-1}{10}.\frac{15}{6}\)

=\(\frac{2}{3}+\frac{-1}{4}\)

=\(\frac{8}{12}+\frac{-3}{12}=\frac{5}{12}\)

bài2

a,\(\left(\frac{2}{7}.x+\frac{3}{7}\right):2\frac{1}{5}-\frac{3}{7}=1\)

=>\(\left(\frac{2}{7}.x+\frac{3}{7}\right):\frac{11}{5}=1+\frac{3}{7}=\frac{10}{7}\)

=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{10}{7}.\frac{11}{5}\)

=>\(\frac{2}{7}.x+\frac{3}{7}=\frac{22}{7}\)

=>\(\frac{2}{7}.x=\frac{22}{7}-\frac{3}{7}=\frac{19}{7}\)

=>\(x=\frac{19}{7}:\frac{2}{7}=\frac{19}{7}.\frac{7}{2}=\frac{19}{2}\)

vậy x\(=\frac{19}{2}\)

a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)

\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)

\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)

\(6^{x-1}.217=6^{15}.217\)

\(6^{x-1}=6^{15}\)

\(x-1=15\)

\(x=16\)

b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)

\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)

\(3^x=3^{13}\)

\(x=13\)

\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)

=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)

=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)

=> \(3^x.3^4-4.3^x=-386339074,3\)

=> \(3^x.\left(3^4-4\right)=-386339074,3\)

=> \(3^x.77=-386339074,3\)

=> \(3^x=-386339074,3:77\)

=> \(3^x=-5017390,575\)

=> x = ... chắc tự ngồi tính đc

cái a tương tự thôi

\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)

\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow27x-2x-4x-27+2=0\)

\(\Leftrightarrow21x=25\)

\(\Leftrightarrow x=\frac{25}{21}\)

Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !

\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)

\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)

\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)

\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)

\(\Leftrightarrow-20x-12=56\)

\(\Leftrightarrow-20x=68\)

\(\Leftrightarrow x=-\frac{17}{5}\)

Tự check lại nhá

\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)

\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)

\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

Ta dễ thấy \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}< 0\) nên

\(x+95=0\Leftrightarrow x=-95\)

a) ĐKXĐ: x khác +2

\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)

<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)

<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)

<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22

<=> x^2 - 7x - 2 = 2x - 22

<=> x^2 - 7x - 2 - 2x + 22 = 0

<=> x^2 - 9x + 20 = 0

<=> (x - 4)(x - 5) = 0

<=> x - 4 = 0 hoặc x - 5 = 0

<=> x = 4 hoặc x = 5

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