4x-2x+x-27:9=33 

2x+x-27:9=33

3x-27:9=33

3x-27=33×9=297

3x=297+27=324

x=324÷3=108

a) \(5^{x-1}+5^{x-3}=650\)

\(\Rightarrow5^x\left(\frac{1}{5}+\frac{1}{125}\right)=650\)

\(\Rightarrow5^x=650:\frac{26}{125}\)

\(\Rightarrow5^x=3125\)

\(\Rightarrow5^x=5^5\)

\(\Rightarrow x=5\)

a) Theo bài ra ta có bảng xét dấu:

-Xét x\(\le\)2

=>-(x-2)+(-x-5)=2x+5

=>-x+2-x+5=2x+5

=>-2x-3=2x+5

=>-5-3=2x+2x

=>-8=4x

=>x=-2 <t/m>

-Xét 2\(\le\)x\(\le\)5

=>(x-2)+(-x-5)=2x+5

=>x-2-x+4=2x+5

=>2=2x+5

=>2-5=2x

=>.-3=2x

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

-Xét x\(\ge\)5

=>(x-2)+(x-5)=2x-5

=>x-2+x-5=2x-5

=>2x-7=2x-5

=>5-7=2x-2x

=>-2=0 <vô lý>

b) Theo bài ra ta có bảng xét dấu:

-Xét x\(\le\)-9

=>(6-x)-(-x+9)=2x+3

=>6-x+x-9=2x+3

=>-3=2x+3

=>-3-3=2x

=>-6=2x

=>x=-3 <ko t/m>

-Xét -9 \(\le\)x\(\le\)6

=>(6-x)-(x+9)=2x+3

=>6-x-x+9=2x+3

=>-2x-3=2x+3

=>-3-3=2x+2x

=>-6=4x

=>x=-6:4=\(\frac{-6}{4}\)=\(\frac{-3}{2}\)<t/m>

-Xét x\(\ge\)6

=>-(6-x)-(x+9)=2x+3

=>-6+x-x+9=2x+3

=>3=2x+3

=>3-3=2x

=>0=2x

=>x=0 <ko t/m>

Nếu thấy đúng thì bấm đúng cho mình nhak

\(a.\left[{}\begin{matrix}2x+9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(b.\left[{}\begin{matrix}x-3=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

+) Xét \(\frac{2x+1}{5}\)= \(\frac{4y-5}{9}\)= \(\frac{2x+4y-4}{7}=0\)

\(\Rightarrow2x+1=0\)

\(\Rightarrow x=\frac{-1}{2}\)

+) Xét \(2x+4y-4\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{2x+1}{5}=\frac{4y-5}{9}=\frac{2x+1+4y-4}{14}=\frac{2x+4y-4}{7x}\)

\(\Rightarrow14=7x\)

\(\Rightarrow x=2\)

Vậy \(x\in\left\{\frac{-1}{2};2\right\}\)

_ Một hằng số làm gì có 3 dấu bằng hả em ? 

Bạn lên https://h.vn/hoi-dap/question/137830.html nha

             Ungr hộ mình nha

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

=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)

=>\(x^2+8x+26-x^2+10x-25=0\)

=>18x+1=0

=>\(x=-\dfrac{1}{18}\)

b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)

=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)

=>\(x^2-28x+43-x^2-6x-9=0\)

=>34-34x=0

=>34x=34

=>x=1

c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)

=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)

=>\(9x^2+17x+4-225-9x^2+225=0\)

=>17x+4=0

=>x=-4/17

a)(2x-3)(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy x=3/2 hoặc x=-5

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

a: Ta có: \(\left(2x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)