38 - x.12 - x.16 = 40

38 - x.(12+16) = 40

38 - 28.x = 40

28.x = 40 - 38

28.x = 2

x= 2/28

x=1/14

\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)

Câu a) xem lại đề giùm nhé em

b) \(\left(x-1\right)^3=9^3\)

\(x-1=9\)

\(x=10\)

Vậy \(x=10\)

c) \(\left(x-1\right)^2=25\)

\(x-1=5\) hoặc \(x-1=-5\)

* \(x-1=5\)

\(x=6\)

* \(x-1=-5\)

\(x=-4\)

Vậy \(x=-4\); \(x=6\)

d) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

Vậy \(x=2\)

e) Sửa đề: \(\left(2x+4\right)^3=64\)

\(\left(2x+4\right)^3=4^3\)

\(2x+4=4\)

\(2x=0\)

\(x=0\)

Vậy \(x=0\)

Sửa đề: 2(x-1)^2+4x-19=(2x-1)(2x+5)

=>2(x^2-2x+1)+4x-19=4x^2+10x-2x-5

=>2x^2-4x+2+4x-19=4x^2+8x-5

=>4x^2+8x-5=2x^2-17

=>2x^2+8x+12=0

=>x^2+4x+6=0

=>(x+2)^2+2=0(loại)

\(\left(2x+3\right)^2=25\)

\(\Rightarrow2x+3=5\) hoặc \(2x+3=-5\)

+) \(2x+3=5\Rightarrow2x=2\Rightarrow x=1\)

+) \(2x+3=-5\Rightarrow2x=-8\Rightarrow x=-4\)

Vậy \(x\in\left\{1;-4\right\}\)

\(\left(2x+3\right)^2=25\)

\(\Rightarrow\left(2x+3\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow2x+3=\pm5\)

* Với \(2x+3=5\)

\(2x=5-3\)

\(2x=2\)

\(x=2\div2\)

\(x=1\)

* Với \(2x+3=-5\)

\(2x=-5-3\)

\(2x=-8\)

\(x=-8\div2\)

\(x=-4\)

Vậy \(x\in\left\{1;-4\right\}\)

2(x² - 18x + 40) = 0 

x² -18x + 40 = 0

x² - (6x + 6x) + 40 = 0

x² - 6x - 6x + 40 = 0 

x ( x - 6 ) - 6 (x - 6) + 36 + 40 = 0 

(x - 6 ) ( x - 6 ) + 76 = 0

( x - 6 )² = -76 ( vô lí vì ( x - 6 )² \(\ge0\forall x\))

Vậy x ko có gtri nào t/m

Bn ơi hình như 18x là bằng 6x.3x chứ

dù gì thì mik vẫn lun thank you bn vì bạn đã giúp mik giải đc 1 phần trong bài toán này

cảm ơn bn nha!!!!!!

HIHI!!!^-^!!!!

a: \(3x\left(x-3\right)+4x-12=0\)

=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)

=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(3x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)

\(\Leftrightarrow x^3+1-x^3+2x=17\)

=>2x+1=17

=>2x=17-1=16

=>\(x=\dfrac{16}{2}=8\)

c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)

=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)

=>\(15x=-14\)

=>\(x=-\dfrac{14}{15}\)

thuylinh1542004

Thái Thị Thùy Linh

a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)

\(2x=144\)

\(x=72\)

b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)

\(\sqrt{9x^2-6x}=9\)

\(9x^2-6x=81\)

\(\left(3x-1\right)^2=82\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)

c) \(x^2\sqrt{5}-\sqrt{125}=0\)

\(x^2\sqrt{5}=5\sqrt{5}\)

\(x^2=5\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

các thầy cô giúp e vs ạ