`B.{x in R|x<=-5/4}`

\(-8x-10=\left|-8x-10\right|\)

\(\left|-8x-10\right|=-8x-10\) khi \(-8x-10\ge0\Leftrightarrow-8x\ge10\Leftrightarrow x\le-\dfrac{5}{4}\)

PT trở thành :

\(-8x-10=-8x-10\)

\(\Leftrightarrow-8x+8x=-10+10\)

\(\Leftrightarrow0x=0\)

PT có vô số nghiệm

Vậy \(S=\){\(x\in R\)| x\(\le-\dfrac{5}{4}\)}

\(\left|-8x-10\right|=-\left(-8x-10\right)=8x+10\) khi \(-8x-10< 0\Leftrightarrow-8x< 10\Leftrightarrow x>-\dfrac{5}{4}\)

PT trở thành :

\(-8x-10=8x+10\)

\(\Leftrightarrow-8x-8x=10+10\)

\(\Leftrightarrow-16x=20\)

\(\Leftrightarrow4x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{4}\)(KTMĐK)

Vậy \(S=\){\(x\in R\)|x \(\le-\dfrac{5}{4}\)}

=> Đáp án B đúng

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

Lời giải:

b.

$\frac{2x}{3}=8$

$\Leftrightarrow 2x=3.8=24$

$\Leftrightarrow x=24:2=12$
d.

$\frac{6}{5}x=-9$

$\Leftrightarrow x=-9: \frac{6}{5}=\frac{-15}{2}$

f.

$\frac{2-3x}{4}=\frac{4x-5}{5}$

$\Leftrightarrow 5(2-3x)=4(4x-5)$

$\Leftrightarrow 10-15x=16x-20$

$\Leftrightarrow 30=31x$

$\Leftrightarrow x=\frac{30}{31}$

h.

$\frac{10-3x}{2}=\frac{6x+1}{3}$

$\Leftrightarrow 3(10-3x)=2(6x+1)$

$\Leftrightarrow 30-9x=12x+2$

$\Leftrightarrow 28=21x$

$\Leftrightarrow x=\frac{28}{21}=\frac{4}{3}$

\(\frac{x}{10}+\frac{x-1}{9}+\frac{x-2}{8}+\frac{x-3}{7}+\frac{x-4}{6}+\frac{x-5}{5}=6\)

=> \(\left(\frac{x}{10}-1\right)+\left(\frac{x-1}{9}-1\right)+\left(\frac{x-2}{8}-1\right)+\left(\frac{x-3}{7}-1\right)+\left(\frac{x-4}{6}-1\right)+\left(\frac{x-5}{5}-1\right)=0\)

=> \(\frac{x-10}{10}+\frac{x-10}{9}+\frac{x-10}{8}+\frac{x-10}{7}+\frac{x-10}{6}+\frac{x-10}{5}=0\)

=> \(\left(x-10\right)\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+\frac{1}{7}+\frac{1}{6}+\frac{1}{5}\right)=0\)

=> x - 10 = 0

=> x = 10

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)

=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)

=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)

=>x=0

b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)

=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)

=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)

Đến đây thì bạn giải giống câu a

bài 1

a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)

d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)

bài 2

a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)

b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)

bài 4 :

145  69 + 22 x  145 +145 x 8 + 145 

\(=145\times\left(69+22+8+1\right)=145\times100=14500\)

bài 1:

a, \(\dfrac{2}{5}+\dfrac{3}{8}=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b,\(\dfrac{7}{6}-\dfrac{2}{3}=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c,\(\dfrac{5}{9}x6=\dfrac{5}{9}x\dfrac{6}{1}=\dfrac{30}{9}\)

d,\(\dfrac{8}{5}:\dfrac{4}{7}=\dfrac{8}{5}x\dfrac{7}{4}=\dfrac{14}{5}\)

bài 2 :

\(a,\dfrac{4}{5}+x=\dfrac{5}{6}\)

            \(x=\dfrac{5}{6}-\dfrac{4}{5}\)

            \(x=\dfrac{1}{30}\)

b, \(x:\dfrac{7}{10}=5\)

    \(x\)         \(=5x\dfrac{7}{10}\)

    \(x\)         \(=\dfrac{35}{10}\)    

bài 3 :

đổi :16 tấn 8 tạ = 168 tạ

        2 tấn 6 tạ = 26 tạ 

xe ô tô thứ nhất chở số tạ hàng là:

      ( 168 + 26 ) : 2= 97 ( tạ)

xe ô tô thứ hai chở số tạ hàng là:

         97 - 26 = 71 ( tạ)

               đáp số :xe ô tô thứ nhất : 97 tạ thóc

                            xe ô tô thứ hai  : 71 tạ thóc 

a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)

=>x+2006=0

=>x=-2006

b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)

=>x-105=0

=>x=105