a) 2x - 3 = -12
=> 2x = -12 + 3 = -9
=> x = \(-\frac{9}{2}\)
b) \(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)
=> \(\frac{1}{2}+2x=-\frac{5}{6}\cdot\frac{3}{2}\)
=> \(\frac{1}{2}+2x=-\frac{5}{2}\cdot\frac{1}{2}\)
=> \(\frac{1}{2}+2x=-\frac{5}{2}\)
=> \(2x=-\frac{5}{2}-\frac{1}{2}=-3\)
=> \(x=-3:2=-\frac{3}{2}\)
c) \(1< \frac{x}{5}< 2\)
=> \(\frac{5}{5}< \frac{x}{5}< \frac{10}{5}\)
=> 5 < x < 10
=> x \(\in\){6,7,8,9}
Dù bạn có cho âm vào nx thì nó vẫn sai nhá
d) Đặt \(A=\frac{x+5}{x-2}=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\)
=> \(x-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
+) x - 2 = 1 => x = 3(T/M)
x - 2 = -1 => x = -1 +2 = 1(t/m)
x - 2 = 7 => x = 9 (t/m)
x - 2 = -7 => x = -7 + 2 = -5(t/m)
e) làm nốt ...
a,\(2x-3=-12\)
\(< =>2x=-12+3=-9\)
\(< =>x=-\frac{9}{2}\)
b,\(\frac{1}{2}+2x=-\frac{5}{6}:\frac{2}{3}\)
\(< =>\frac{1}{2}+\frac{4x}{2}=-\frac{5}{6}.\frac{3}{2}\)\(< =>\frac{4x+1}{2}=-\frac{5}{4}\)
\(< =>\frac{8x+2}{4}=-\frac{5}{4}\)\(< =>8x+2=-5\)
\(< =>8x=-5-2=-7\)\(< =>x=-\frac{7}{8}\)
Ây da,câu b nhầm tận đáy :))
Bạn tham khảo câu b dcv_new nhé
Mình làm nốt câu cuối
\(\frac{2}{3x}-\frac{3}{2\left(x-\frac{1}{2}\right)}=\frac{5}{12}\left(đkxđ:x\ne0;\frac{1}{2}\right)\)
\(< =>\frac{2}{3x}-\frac{3}{2x-1}=\frac{5}{12}\)
\(< =>\frac{2\left(2x-1\right)}{3x\left(2x-1\right)}-\frac{3\left(3x\right)}{\left(2x-1\right)3x}=\frac{5}{12}\)
\(< =>\frac{4x-2}{6x^2-3x}-\frac{9x}{6x^2-3x}=\frac{5}{12}\)
\(< =>\frac{-5x-2}{6x^2-3x}=\frac{5}{12}\)
\(< =>\left(-5x-2\right).12=\left(6x^2-3x\right).5\)
\(< =>-60x-24=30x^2-15x\)
\(< =>30x^2-15x+60x+24=0\)
\(< =>30x^2+45x+24=0\)
Ta có : \(\Delta'=b'^2-ac=\left(\frac{45}{2}\right)^2-30.24=-\frac{855}{4}\)
Vì \(\Delta'< 0\)nên phương trình vô nghiệm
Vậy không có x thỏa mãn đề bài
