(x-26)⋅2=4^2
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321 x 3 + 321 x 5 + 321 x 2
= 321 x (3 + 5 + 2)
= 321 x 10
= 3 210.
321 x 3 + 321 x 5 + 321 x 2
= 321 x (3 + 5 + 2)
= 321 x 10
= 3210

\(C,\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{2}.\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{2}.\dfrac{4}{4}=\dfrac{1}{2}.1=\dfrac{1}{2}\)
\(D,\dfrac{11}{3}.\dfrac{26}{7}-\dfrac{26}{7}.\dfrac{8}{3}=\dfrac{26}{7}.\left(\dfrac{11}{3}-\dfrac{8}{3}\right)\)
\(=\dfrac{26}{7}.\dfrac{3}{3}=\dfrac{26}{7}.1=\dfrac{26}{7}\)

\(ĐK:x\ne\pm2\)
\(\Rightarrow\dfrac{\left(x+2\right)\left(x+2\right)-\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-26}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-26\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4+26=0\)
\(\Leftrightarrow8x+26=0\)
\(\Leftrightarrow x=-\dfrac{26}{8}\left(tm\right)\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+2^2\right)-x\left(x^2-3^2\right)=26\\ \Leftrightarrow x^3+2^3-x^3+9x=26\\ \Leftrightarrow9x+8=26\\ \Rightarrow x=\dfrac{26-8}{9}=2\)
Lời giải:
$(x+2)(x^2-2x+4)-x(x+3)(x-3)=26$
$\Leftrightarrow x^3+8-x(x^2-9)=26$
$\Leftrightarrow 9x+8=26$
$\Leftrightarrow 9x=18$
$\Leftrightarrow x=2$


Ta có: (x + 2)(x + 4) > (x – 2)(x + 8) + 26
⇔ x 2 + 6x + 8 > x 2 + 6x + 10
⇔ x 2 + 6x - x 2 - 6x > 10 - 8
⇔ 0x > 2
Vậy bất phương trình vô nghiệm.

\(=\dfrac{3}{2}\cdot\dfrac{4}{2}\cdot\dfrac{5}{3}\cdot\dfrac{6}{4}\cdot...\cdot\dfrac{27}{25}\cdot\dfrac{28}{26}\cdot\dfrac{29}{27}\)
\(=\dfrac{1}{4}\cdot28\cdot29=7\cdot29=203\)
(x-26).2= 16
x-26 = 16:2
x-26 = 8
x = 8+26
x = 34
Vậy x=34
giup to voi