1: \(x^2-5x+1=0\)

\(\Delta=\left(-5\right)^2-4\cdot1\cdot1=25-4=21>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{5-\sqrt{21}}{2\cdot1}=\frac{5-\sqrt{21}}{2}\\ x=\frac{5+\sqrt{21}}{2\cdot1}=\frac{5+\sqrt{21}}{2}\end{array}\right.\)

2: \(3x^2-2x+5=0\)

\(\Delta=\left(-2\right)^2-4\cdot3\cdot5=4-4\cdot15=4-60=-56<0\)

=>Phương trình vô nghiệm

3: \(3x^2+2x+5=0\)

\(\Delta=2^2-4\cdot3\cdot5=4-4\cdot15=4-60=-56<0\)

=>Phương trình vô nghiệm
4: \(3x^2-4x+1=0\)

=>\(3x^2-3x-x+1=0\)

=>3x(x-1)-(x-1)=0

=>(x-1)(3x-1)=0

=>\(\left[\begin{array}{l}x-1=0\\ 3x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=\frac13\end{array}\right.\)

Bài có khúc bị khuyết em nha! Mà lại khúc quan trọng nữa

Khúc nào chị, chị thu gọn á, nó sẽ nổi ra hết

a) \(M_X=M_{Br2}=160\) (đvC)

b) CT của hợp chất : X2O3

Ta có : \(2X+16.3=160\)

=> X=56

Vậy X là Fe

k: \(\left(4x-16\right)\left(-72+9x\right)=0\)

=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)

=>\(36\left(x-4\right)\left(x-8\right)=0\)

=>\(\left(x-4\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)

m: \(\left(20+5x\right)\left(4x-8\right)=0\)

=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)

=>\(\left(x+4\right)\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

n: \(\left(-4x+48\right)\left(2x-24\right)=0\)

=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)

=>\(\left(x-12\right)^2=0\)

=>x-12=0

=>x=12

o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)

=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)

=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)

p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)

=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)

=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)

=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)

q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)

=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)

=>\(x\left(x-2\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)

r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)

=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)

=>\(x\left(x+3\right)\left(x-8\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)

s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)

=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)

=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)

t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)

=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)

=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)

ΔABC vuông tại A
=>\(AB^2+AC^2=BC^2\)

=>\(AB^2=10^2-6^2=100-36=64=8^2\)

=>AB=8(cm)

BE+EC=BC

=>CE=10-4=6(cm)

Xét ΔCAB có HE//AB

nên \(\frac{HE}{AB}=\frac{CH}{CA}=\frac{CE}{CB}\)

=>\(\frac{HE}{8}=\frac{CH}{6}=\frac{6}{10}=\frac35\)

=>\(HE=8\cdot\frac35=4,8\left(\operatorname{cm}\right);CH=6\cdot\frac35=3,6\left(\operatorname{cm}\right)\)

CH+HA=AC

=>AH=6-3,4=2,4(cm)

ΔABH vuông tại A

=>\(BH^2=AB^2+AH^2\)

=>\(BH^2=8^2+2,4^2=64+5,76=69,76\)

=>\(BH=\sqrt{69,76}\left(\operatorname{cm}\right)\)

\(g,ĐK:x\ge0\\ PT\Leftrightarrow10\sqrt{x}+8\sqrt{x}-11\sqrt{x}=21\\ \Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\\ h,ĐK:x\ge0\\ PT\Leftrightarrow6\sqrt{3x}+2\sqrt{3x}-3\sqrt{3x}=15\\ \Leftrightarrow\sqrt{3x}=5\Leftrightarrow3x=25\Leftrightarrow x=\dfrac{25}{3}\left(tm\right)\\ i,ĐK:x\ge0\\ PT\Leftrightarrow12\sqrt{x}-21-2\sqrt{x}+10=6\sqrt{x}-12\\ \Leftrightarrow4\sqrt{x}=-1\Leftrightarrow\sqrt{x}=-\dfrac{1}{4}\Leftrightarrow x\in\varnothing\\ j,ĐK:x\ge2\\ PT\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{1}{5}\sqrt{x-2}=20+4\sqrt{x-2}\\ \Leftrightarrow\sqrt{x-2}=-20\Leftrightarrow x\in\varnothing\)

\(k,ĐK:x\ge3\\ PT\Leftrightarrow6\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\Leftrightarrow x=28\left(tm\right)\\ l,ĐK:x\ge5\\ PT\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

Bạn có thể giải một cách chi tiết giúp mình đc k ạ huhu