`#040911`

a,

\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)

Vậy, \(x=-\dfrac{8}{21}\)

b,

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy, \(x\in\left\{-2;3\right\}\)

c,

\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)

Bạn xem lại đề có sai kh nhỉ?

c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)

Đề bài sai, pt này ko giải được

Đề đúng: \(\dfrac{4x^2+16}{x^2+6}=...\)

Mẫu số bên trái thừa mất số 1

b) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)

\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)

c) \(\left(3x+2\right)^3=-27\)

\(\Rightarrow\left(3x+2\right)^3=\left(-3\right)^3\)

\(\Rightarrow3x+2=-3\)

\(\Rightarrow3x=\left(-3\right)-2\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=\left(-5\right):3\)

\(\Rightarrow x=-\frac{5}{3}\)

Vậy \(x=-\frac{5}{3}.\)

Chúc bạn học tốt!

Bạn ơi, gõ Công thức trực quan cho dễ nhìn đi bạn! :)

a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)

\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)

\(\Rightarrow x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}.\)

d) \(27^x:3^x=9\)

\(\Rightarrow\left(27:3\right)^x=9\)

\(\Rightarrow9^x=9\)

\(\Rightarrow9^x=9^1\)

\(\Rightarrow x=1\)

Vậy \(x=1.\)

e) \(\frac{16}{2^x}=2\)

\(\Rightarrow2^x=16:2\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

Bài 2: 

a: Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)

\(=\left(x^2-4\right)\left(x^2+4\right)\)

\(=x^4-16\)

b: Ta có:\(\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3-x^2y+xy^2+x^2y-xy^2+y^3\)

\(=x^3+y^3\)

Bài 1: 

Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+3\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x\left(x^2+4x+3\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x^3-4x^2-3x+3x^2=0\)

\(\Leftrightarrow-x^2-3x+64=0\)

\(\Leftrightarrow x^2+3x-64=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-64\right)=265\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{265}}{2}\\x_2=\dfrac{-3+\sqrt{265}}{2}\end{matrix}\right.\)

\(a,\frac{1}{2}x+\frac{2}{3}\left(x-2\right)=\frac{1}{3}\)

\(\frac{1}{2}x+\frac{2}{3}x-\frac{4}{3}=\frac{1}{3}\)

\(\frac{7}{6}x=\frac{5}{3}\)

\(x=\frac{10}{7}\)

\(b,16^{x-1}:16=4^4\)

\(16^{x-1}=4096\)

\(16^{x-1}=16^3\)

\(\Rightarrow x-1=3\)

\(x=4\)

=.= hk tốt!!

a) \(\frac{1}{2}x+\frac{2}{3}\left(x-2\right)=\frac{1}{3}\)

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

b)16^x-1:16=256 => 16^x-1=4096=16³

^{T thấy x-1=3 =>x=2}

Chúc bạn học tốt

a) x = 3

b) x = 2

c) x = 4

d) x = 1.