c: ta có: \(7x^2-2x-5=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{7}\end{matrix}\right.\)

\(x\inƯC\left(120;150\right)\)

mà x>12

nên \(x\in\left\{15;30\right\}\)

\(\left|x+1\right|=x\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+1=x\)

\(\Leftrightarrow1=0\left(VLý\right)\)

Vậy \(S=\varnothing\)

\(|x+1|=x\) (đk: \(x\ge-1\))

<=> \(\left[{}\begin{matrix}x+1=x\\x+1=-x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}0=-1\left(VLí\right)\\x=\dfrac{-1}{2}\left(TM\right)\end{matrix}\right.\)

Vậy nghiệm PT là \(S=\left\{\dfrac{-1}{2}\right\}\)

\(\frac{x}{5}=\frac{20}{x}\)

Ta có :

x . x = 20 . 5

x² = 100

=> x = 10 hoặc x = -10

XxX=20x5

X=100

10

tương đương vỡi^2=100. tương đương với x=10

=5²⁰:(5¹⁵*(6+19))

=5²⁰:(5¹⁵*25)

=5²⁰:(5¹⁵*5²)

=5²⁰:5¹⁷=5³=125

k cho mình nha

\(5^{20}:\left(5^{15}.6+5^{15}.19\right)\)

=\(5^{20}:5^{15}.\left(6+19\right)\)

=\(5^{20}:5^{15}.25\)

=\(5^{20}:5^{15}.5^2\)

=\(5^{20}:5^{17}\)

=\(5^3\)=125.

Đáp án bằng : 125

(4,53 x 02+6,165 x 3):2,5

=(0,906  + 18,495); 2,5

=19,401: 2,5

=7,7604

a: \(x⋮10\)

b: \(x⋮5\)

x=80 vì 80 đạt đủ mọi tiêu chẩn

Đề :

\(199,5:\frac{\frac{3}{5}\cdot x+4}{20}+8,5=\frac{242}{5}\)

\(\Rightarrow199,5:\frac{\frac{3}{5}\cdot x+4}{20}=\frac{242}{5}-8,5\)

\(\Rightarrow199,5:\frac{\frac{3}{5}\cdot x+4}{20}=48,4-8,5\)

\(\Rightarrow199,5:\frac{\frac{3}{5}\cdot x+4}{20}=39,9\)

\(\Rightarrow\frac{\frac{3}{5}\cdot x+4}{20}=199,5:39,9\)

\(\Rightarrow\frac{\frac{3}{5}\cdot x+4}{20}=5\)

\(\Rightarrow\frac{3}{5}\cdot x+4=5\cdot20\)

\(\Rightarrow\frac{3}{5}\cdot x+4=100\)

\(\Rightarrow\frac{3}{5}x=96\Leftrightarrow x=160\)

biến đổi là ra x = 160 thôi

la sao bn