\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)

\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)

\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{25}{28}\)

\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)

\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)

\(\Rightarrow x=-\dfrac{37}{30}\)

\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)

\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)

\(\Rightarrow35x=-70\)

\(\Rightarrow x=-2\)

\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)

\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)

\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)

\(\Rightarrow x=-\dfrac{8}{9}\)

`a)x-3/4=1/7`

`=>x=1/7+3/4`

`=>x=4/28+21/28`

`=>x=25/28`

`b)x+7/5=9/8 . 4/27`

`=>x+7/5=(9.4)/(8.27)`

`=>x+7/5=1/6`

`=>x=1/6+7/5`

`=>x=5/30+42/30`

`=>x=47/30`

`c)2/5-3/7=x/70`

`=>14/35-15/35=x/70`

`=>-1/35=x/70`

`=>-2/70=x/70`

`=>x=-2`

`d)2/9-7/8x=1`

`=>7/8x=2/9-1`

`=>7/8x=-7/9`

`=>x=-7/9:7/8`

`=>x=-7/9 . 8/7`

`=>x=-8/9`

`#lv`

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5

\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)

Bài 2:

a) \(\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{x-2}-5\sqrt{x-1}=7\)

\(\Leftrightarrow0=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(\sqrt{2x^2-3}=4\left(đk:-\sqrt{\dfrac{3}{2}}\ge x\ge\sqrt{\dfrac{3}{2}}\right)\)

\(\Leftrightarrow2x^2-3=16\)

\(\Leftrightarrow2x^2=19\Leftrightarrow x^2=\dfrac{19}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

`C(x)=`\(5-8x^4+2x^3+x+5x^4+x^2-4x^3\)

`C(x)= (-8x^4+5x^4)+(2x^3-4x^3)+x^2+x+5`

`C(x)= -3x^4-2x^3+x^2+x+5`

`D(x)=`\(\left(3x^5+x^4-4x\right)-\left(4x^3-7+2x^4+3x^5\right)\)

`D(x)= 3x^5+x^4-4x-4x^3+7-2x^4-3x^5`

`D(x)=(3x^5-3x^5)+(x^4-2x^4)-4x^3-4x+7`

`D(x)=-x^4-4x^3-4x+7`

`P(x)=C(x)+D(x)`

`P(x)=( -3x^4-2x^3+x^2+x+5)+(-x^4-4x^3-4x+7)`

`P(x)=-3x^4-2x^3+x^2+x+5-x^4-4x^3-4x+7`

`P(x)=(-3x^4-x^4)+(-2x^3-4x^3)+x^2+(x-4x)+(5+7)`

`P(x)=-4x^4-6x^3+x^2-3x+12`

`Q(x)=C(x)-D(x)`

`Q(x)=( -3x^4-2x^3+x^2+x+5)-(-x^4-4x^3-4x+7)`

`Q(x)=-3x^4-2x^3+x^2+x+5+x^4+4x^3+4x-7`

`Q(x)=(-3x^4+x^4)+(-2x^3+4x^3)+x^2+(x+4x)+(5-7)`

`Q(x)=-2x^4+2x^3+x^2+5x-2`

`F(x)=Q(x)-(-2x^4+2x^3+x^2-12)`

`F(x)=(-2x^4+2x^3+x^2+5x-2)-(-2x^4+2x^3+x^2-12)`

`F(x)=-2x^4+2x^3+x^2+5x-2+2x^4-2x^3-x^2+12`

`F(x)=(-2x^4+2x^4)+(2x^3-2x^3)+(x^2-x^2)+5x+(-2+12)`

`F(x)=5x+10`

Đặt `5x+10=0`

`\Leftrightarrow 5x=0-10`

`\Leftrightarrow 5x=-10`

`\Leftrightarrow x=-10 \div 5`

`\Leftrightarrow x=-2`

Vậy, nghiệm của đa thức là `x=-2.`

\(a,\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{-1}{7}\\ \dfrac{3}{2}x=\dfrac{-1}{7}-\dfrac{1}{4}\\ \dfrac{3}{2}x=-\dfrac{11}{28}\\ x=\dfrac{-11}{28}:\dfrac{3}{2}\\ x=-\dfrac{11}{42}\\ b,x:\dfrac{4}{7}-\dfrac{1}{7}=0,5\\ x:\dfrac{4}{7}-\dfrac{1}{7}=\dfrac{1}{2}\\ x:\dfrac{4}{7}=\dfrac{1}{2}+\dfrac{1}{7}\\ x:\dfrac{4}{7}=\dfrac{9}{14}\\ x=\dfrac{9}{14}\times\dfrac{4}{7}\\ x=\dfrac{18}{49}\)

VD: 3/2 có nghĩa là 3 phần 2 nha làm vậy cho mn dễ hiểu. Mọi người theo dõi mik, để có bất ngờ nha. Chúc mọi người 1 ngày vui vẻ, ngày càng học giỏi nhé!

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)

\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)

Vậy A = \(\frac{1}{2}\)

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)

Vậy B = \(\frac{1}{2}\)