a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20

10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110

10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110

10 x X - 100 = 110

10 x X = 110 + 100

10 x X = 210

       X = 210 : 10

       X = 21

a 10 x X-1-3-5-7-....-19 = 2+4+6+....+20

​10xX-1-3-5-7-....-19=110

​10xX=110+1+3+5+7+....+19

​10xX=210

​X=210:10

​X=21

b là 4

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\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

a. x/2-3x/5+13/5=-7/5-7/10x

    -7/5-x/2+3x/5-13/5=7/10x

    -x/2+3x/5-4=7/10x

    7/10x-3x/5+x/2=-4

    7x-6x+5x/10=-4

     6x=-40

     x=-20/3

\(a,\frac{-3}{2}-2x+\frac{3}{4}=-1\)

\(\frac{-3}{2}-2x=-1-\frac{3}{4}\)

\(\frac{-3}{2}-2x=\frac{-7}{4}\)

\(2x=\frac{-7}{4}+\frac{-3}{2}\)

\(2x=\frac{-13}{4}\)

\(x=\frac{-13}{4}:2\)

\(x=\frac{-13}{4}.\frac{1}{2}\)

\(x=\frac{-13}{8}\)

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

<=> \(2\left(x+3\right)-x\left(x+3\right)=0\)

<=> \(\left(x+3\right)\left(2-x\right)=0\)

<=> \(\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

\(2x\left(3x-5\right)=10-6x\)

<=> \(2x\left(3x-5\right)-\left(10-6x\right)=0\)

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

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

<=> \(2\left(3x-5\right)\left(x+1\right)=0\)

<=> \(\orbr{\begin{cases}3x-5=0\\x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

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

\(\Leftrightarrow\)\(2x+6-x^2-3x=0\)

\(\Leftrightarrow\)\(-x^2-x+6=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy....

        \(2x\left(3x-5\right)=10-6x\)

\(\Leftrightarrow\)\(6x^2-10x=10-6x\)

\(\Leftrightarrow\)\(6x^2-4x-10=0\)

\(\Leftrightarrow\)\(2\left(x+1\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\3x-5=0\end{cases}}\)

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

Vậy....

a) \(x=\dfrac{25}{72}\)

b)\(x=-\dfrac{1}{4}\)

  \(x=\dfrac{3}{2}\)

c)\(x=\dfrac{5}{4}\) hoặc

  x \(=\dfrac{8}{5}\)

d và e chịu vì mk kg giỏi lắm về mũ 

f)\(x=-2\)

G)\(x=-\dfrac{5}{12}\)

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)