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\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=\left(x+14\right)-\left(x+2\right)\)
\(\Leftrightarrow x=x+14-x-2\)
\(\Leftrightarrow x=\left(x-x\right)+\left(14-2\right)\)
\(\Leftrightarrow x=0+12\)
\(\Leftrightarrow x=12\)
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)
\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)
\(10x-70-8x-40=-6\)
\(10x-8x=\left(-6\right)+70+40\)
\(2x=104\)
\(x=104\div2\)
\(x=52\)
b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)
\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)
\(8x-16-21-7x=6\)
\(8x-7x=6+16+21\)
\(x=43\)
=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{1}{16}\)
=> \(\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{1}{16}\)
=> \(\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right).\left(x+14\right)}=\frac{1}{16}\)
=> \(\frac{x+14-x-2}{x\left(2+14\right)}=\frac{1}{16}\)
=> \(\frac{12}{16x}=\frac{1}{16}\)
=> x = 12
toán lớp 6 nâng cao bạn ạ