ĐKXĐ: x∉{2/3;4}

Ta có: \(\frac{14}{3x-2}-\frac{x+2}{x-4}=\frac{3}{8-2x}-\frac56\)

=>\(\frac{14}{3x-2}-\frac{x+2}{x-4}-\frac{3}{8-2x}=-\frac56\)

=>\(\frac{14}{3x-2}-\frac{x+2}{x-4}+\frac{3}{2\left(x-4\right)}=-\frac56\)

=>\(\frac{14}{3x-2}-\frac{2x+4}{2\left(x-4\right)}+\frac{3}{2\left(x-4\right)}=-\frac56\)

=>\(\frac{14}{3x-2}+\frac{-2x-4+3}{2\left(x-4\right)}=-\frac56\)

=>\(\frac{14}{3x-2}+\frac{-2x-1}{2\left(x-4\right)}=-\frac56\)

=>\(\frac{14\cdot2\cdot\left(x-4\right)+\left(-2x-1\right)\left(3x-2\right)}{2\left(x-4\right)\left(3x-2\right)}=\frac{-5}{6}\)

=>\(\frac{28x-112-6x^2+4x-3x+2}{2\left(x-4\right)\left(3x-2\right)}=-\frac56\)

=>\(\frac{-6x^2+29x-110}{2\left(3x^2-14x+8\right)}=\frac{-5}{6}\)

=>\(6\left(-6x^2+29x-110\right)=-5\cdot2\cdot\left(3x^2-14x+8\right)\)

=>\(-36x^2+174x-660=-30x^2+140x-80\)

=>\(-6x^2+34x-580=0\)

=>\(3x^2-17x+290=0\)

\(\Delta=\left(-17\right)^2-4\cdot3\cdot290=289-12\cdot290<0\)

=>Phương trình vô nghiệm

thank you very much

a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)

Phần b T²

c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)

cảm ơn bạn nha

\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)

nhầm rùi các bạn ạ có cả x =12

uh đúng r

Đáp án lần lượt là: \(\frac{23}{36};\frac{31}{7};\frac{1}{6};\frac{3}{5}\)

#Hk_tốt

#Ken'z

NGọc Ken nhanh nhất

a) $\frac{4}{{25}}:\frac{4}{3} = \frac{4}{{25}} \times \frac{3}{4} = \frac{3}{{25}}$ 

b) $\frac{3}{{14}}:\frac{6}{7} = \frac{3}{{14}} \times \frac{7}{6} = \frac{{3 \times 7}}{{14 \times 6}} = \frac{{3 \times 7}}{{7 \times 2 \times 3 \times 2}} = \frac{1}{4}$

c) $\frac{{12}}{{15}}:2 = \frac{{12}}{{15}} \times \frac{1}{2} = \frac{{12 \times 1}}{{15 \times 2}} = \frac{{6 \times 2 \times 1}}{{15 \times 2}} = \frac{6}{{15}}$           

d) $\frac{{21}}{8}:6 = \frac{{21}}{8} \times \frac{1}{6} = \frac{{21 \times 1}}{{8 \times 6}} = \frac{{7 \times 3 \times 1}}{{8 \times 3 \times 2}} = \frac{7}{{16}}$

b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)

\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)

\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)

\(=\frac{22}{29}.\)

Chúc bạn học tốt!

a)\(\frac{5}{21}\)+\(\frac{-3}{7}\)<\(\frac{x}{21}\)<\(\frac{-2}{7}\)+\(\frac{8}{21}\)

\(\Rightarrow\)\(\frac{-4}{21}\)<\(\frac{x}{21}\)<\(\frac{2}{21}\)

\(\Rightarrow\)\(\frac{x}{21}\)\(\in\)\(\left\{\frac{-3}{21};\frac{-2}{21};\frac{-1}{21};\frac{0}{21};\frac{1}{21}\right\}\)

vậy x\(\in\)\(\left\{-3;-2;-1;0;1\right\}\)

mấy câu kia cs tương tự ạ