no bé ơi

Lời giải:

$\frac{1+14x}{5}=\frac{5-3x}{4}$

$\Rightarrow 4(1+14x)=5(5-3x)$

$4+56x=25-15x$

$56x+15x=25-4$

$71x=21$

$x=\frac{21}{71}$

c, \(\Rightarrow4+56x=25-15x\Leftrightarrow71x=21\Leftrightarrow x=\dfrac{21}{71}\)

số hs khá là: \(\dfrac{3}{4}\times\dfrac{1}{6}=\dfrac{1}{8}\left(hs\right)\)

a: số hs lớp 6A là : \(\dfrac{3}{4}+\dfrac{1}{8}+5=\dfrac{18}{24}+\dfrac{3}{24}+5=5\dfrac{21}{24}=5\dfrac{7}{8}=\dfrac{47}{8}\left(hs\right)\)

b: tỉ số là : \(\dfrac{3}{4}\times100:\dfrac{1}{8}=\dfrac{300}{4}:\dfrac{1}{8}=\dfrac{300\times8}{4\times1}=600\%\)

Gọi học sinh lớp 6a là x hs

Số học sinh giỏi là : \(\dfrac{3}{4}x\)(hs)

Số học sinh khác là : \(\dfrac{1}{6}.\dfrac{3}{4}x=\dfrac{1}{8}x\) (hs)

Theo bài ra có :

\(x-\dfrac{3}{4}x-\dfrac{1}{8}x=5\)

\(x\left(1-\dfrac{3}{4}-\dfrac{1}{8}\right)=5\)

\(\dfrac{1}{8}x=5\)

\(x=40\)

Vậy số hcj sinh lớp 6a là 40 hs

Số học sinh giỏi là : \(\dfrac{3}{4}.40=30\) (hs)

Số học sinh khá là :\(\dfrac{1}{6}.30=5\) (hs)

ĐS ...

\(b,\Leftrightarrow\left\{{}\begin{matrix}m+1=3\\m-3\ne-3\end{matrix}\right.\Leftrightarrow m=2\\ c,\text{PT giao Ox tại hoành độ 3: }\\ x=-3;y=0\Leftrightarrow\left(m+1\right)\left(-3\right)+m-3=0\\ \Leftrightarrow-2m-6=0\Leftrightarrow m=-3\)

a.

ĐKXĐ: \(\left[{}\begin{matrix}x\ge-1+\sqrt{2}\\x\le-1-\sqrt{2}\end{matrix}\right.\)

\(x^2-2x-1+2\left(x-1\right)\sqrt{x^2+2x-1}=0\)

\(\Leftrightarrow\left(x^2+2x-1\right)+2\left(x-1\right)\sqrt{x^2+2x-1}-4x=0\)

\(\Delta'=\left(x-1\right)^2+4x=\left(x+1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=1-x+x+1\\\sqrt{x^2+2x-1}=1-x-x-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=2\\\sqrt{x^2+2x-1}=-2x\left(x\le0\right)\end{matrix}\right.\)

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

\(\Rightarrow x\)

b.

ĐKXĐ: \(x\ge-\sqrt[3]{3}\)

\(x^3+3-\left(5x-1\right)\sqrt{x^3+3}+6x^2-2x=0\)

Đặt \(\sqrt{x^3+3}=t\ge0\)

\(\Rightarrow t^2-\left(5x-1\right)t+6x^2-2x=0\)

\(\Delta=\left(5x-1\right)^2-4\left(6x^2-2x\right)=\left(x-1\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^3+3}=2x\left(x\ge0\right)\\\sqrt{x^3+3}=3x-1\left(x\ge\dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^3+2=4x^2\left(x\ge0\right)\\x^3+3=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x^2-3x-3\right)=0\left(x\ge0\right)\\\left(x-1\right)\left(x^2-8x-2\right)=0\left(x\ge\dfrac{1}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)

\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)

\(\Leftrightarrow2x+8=-x+11\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Nhân chéo ta được\(-6(x+4)=3(x-11)=>-6x-24=3x-33=>6x-3x-24+33=0=>3x+9=0=>3x=-9=>x=-3\)

cái này thì triệt tiêu ra đó bn

là bằng 2020/6069

A = 1/3.5.7 + 1/5.7.9+ ... + 1/2019.2021.2023

2A = 2/3.5.7 + 2/5.7.9+ ... + 2/2019.2021.2023

2A = 1/3-1/5+1/7+1/5-1/7+1/9+....+1/2019-1/2021+1/2023

2A = 1/3 - 1/2023

2A = 2023/6069 - 3/6069

2A = 2023-3/6069

2A = 2020/6069

 A = 1010/6069

Vậy A = 1010/6069

\(A=\sqrt{\dfrac{3\sqrt{5}-1}{3+2\sqrt{5}}}-\sqrt{\dfrac{\sqrt{5}-1}{\sqrt{5}-2}}\)

\(=\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}\)

\(=-\sqrt{2}\)