1,ta có:h(x) = ( x - 3 ).( 16 - 4x )=0

*)x-3=0

=>x=3

*)16-4x=0

=>4x=16

=>x=4

2,ta có:4x^2 - 6x=0

<=>2x(2x-3)=0

*)2x=0

=>x=0

*)2x-3=0

=>2x=3

=>x=\(\frac{3}{2}\)

3,ta có:x^2 + 7x - 8=0

denta:7²-(-4(1.8))=81

x₁:(-7+9):2=1

x₂:(-7-81):2=-8

câu 2 you chả giống kqua tôi gì

1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)

Vậy ...................

b/ ĐKXĐ:\(x\ne2;x\ne5\)

.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x^2-10x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)

Vậy ..............

`Answer:`

`1.`

a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)

b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)

\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)

`2.`

\(ĐKXĐ:x\ne-m-2;x\ne m-2\)

Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)

a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)

b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì

\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)

a) \(\left(x-\frac{1}{3}\right)^2-\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{1}{4}\)

\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\left(\frac{1}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{-1}{6}\end{cases}}}\)

Vậy x= 5/6 hoặc -1/6

b) - Nếu x=0 thì \(5^y=2^0+624=1+624=625=5^4\Rightarrow y=4\left(y\in N\right)\)

- Nếu x \(\ne\) 0 thì vế trái là số chẵn , vế phải là số lẻ \(\forall x;y\inℕ\) ( vô lí)

Vậy x=0, y=4

thank you bạn.

\(x^2-2\left(m+1\right)x+3m-3=0\left(1\right)\)

\(\Delta'>0\Leftrightarrow\left(m+1\right)^2-\left(3m-3\right)=m^2-m+4>0\left(đúng\forall m\right)\)

\(đk\) \(tồn\) \(tại:\sqrt{x1-1}+\sqrt{x2-1}\)

\(\Leftrightarrow1\le x1< x2\Leftrightarrow\left\{{}\begin{matrix}\left(x1-1\right)\left(x2-1\right)\ge0\\x1+x2-2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x1x2-\left(x1+x2\right)+1\ge0\\2\left(m+1\right)-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3m-2-2\left(m+1\right)+1\ge0\\m>0\end{matrix}\right.\)

\(\Leftrightarrow m\ge4\)

\(\Rightarrow\sqrt{x1-1}+\sqrt{x2-1}=4\Leftrightarrow x1+x2-2+2\sqrt{\left(x1-1\right)\left(x2-1\right)}=16\)

\(\Leftrightarrow2\left(m+1\right)+2\sqrt{x1.x2-\left(x1+x2\right)+1}=18\)

\(\Leftrightarrow\left(m+1\right)+\sqrt{3m-3-2\left(m+1\right)+1}=9\)

\(\Leftrightarrow m-4+\sqrt{m-4}=4\)

\(đặt:\sqrt{m-4}=t\ge0\Rightarrow t^2+t=4\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{17}}{21}\left(tm\right)\\t=\dfrac{-1-\sqrt{17}}{21}\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{m-4}=\dfrac{-1+\sqrt{17}}{21}\Leftrightarrow m=....\)

\(\)

cần gấp lắm ạ xin các bạn giúp đỡ . ___ .

Câu 1: 

0,9 x 218 x 2 + 0,18 x 4290 + 0,6 x 353 x 3

= 9/10 x 436 + 9/50 x 4290 + 6/10 x 1059

= 9 x 43,6 + 9 x 85,8 + 6 x 105,9

= 3 x 130,8 + 3 x 257,4 + 3 x 211,8

= 3 x ( 130,8 + 257,4 + 211,8 )

= 3 x 600 

= 1800

Câu 2: 

3/4 x X + 1/2 x X - 15 = 35

X x ( 3/4 + 1/2 ) - 15 = 35

X x ( 3/4 + 1/2 ) = 50

X x 5/4 = 50

X = 40

VẬy X = 40

a) \(1+2+3+...+x=325\)

\(\dfrac{x\times\left(x+1\right)}{2}=325\)

\(x\times\left(x+1\right)=325\times2\)

\(x\times\left(x+1\right)=650\)

\(x\times\left(x+1\right)=25\times26\)

\(x=25\)

b) Bạn xem lại đề: