\(a,5x-5^2=3x+25\)

\(5x-3x=25+5^2\)

\(2x=50\)

\(x=50:2\)

\(x=25\)

\(b,3x+69:23=2x-108:36\)

\(3x+3=2x-3\)

\(3x-2x=-3-3\)

\(x=-6\)

Học tốt

a, 5x - 5² = 3x + 25

=> 5x - 25 = 3x + 25

=> 5x - 3x = 25 + 25

=> 2x = 50

=> x = 25

b, 3x + 69 : 23 = 2x - 108 : 36

=> 3x + 3 = 2x - 3

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

=> x = - 6

a,
5x-5^2=3x+25
5x-25=3x+25
2x=50 => x = 25
b,
3x+3=2x-3
3x-2x=-6 => x=-6

Câu 1:

a) 2(x-3)-3(x-5)=4(3-x)-18

<=> 3x-6-3x+15-12+4x+18=0

<=> 4x+15=0

<=> 4x=-15

<=> x=-15/4

b) -2(2x-8)+3(4-2x)=-57-5(3x-7)

<=> -4x+16+12-6x+57+15x-35=0

<=> -5x+50=0

<=> -5x=-50

<=> x=10

c) 3|2x²-7|=33

<=> |2x²-7|=11

<=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=9\\x^2=-2\end{cases}\Leftrightarrow}x=\pm3}\)

d) có 9x+17=3(3x+2)+11

=> 11 chia hết cho 3x+2

=> 3x+2 thuộc Ư (11)={-11;-1;1;11}

ta có bảng

Câu 2:

xy-5x+y=17

<=> x(y-5)+(y-5)=12

<=> (y-5)(x+5)=12

=> y-5; x+5 \(\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

lập bảng tương tự câu 1

Câu 1

a,b,sorry bạn mk ko bít

c,3.|2x² - 7|=33

=>|2x² - 7|=11

=>2x² -7=11 hoặc 2x² - 7= -11

=>2x² =18 hoặc 2x²= -4

=>x²=9 hoặc x²= -2

=>x²=3² 

Vậy x=3

d,9x+17 chia hết cho 3x+2

=>3(3x+2)-6+17 chia hết cho 3x+2

=>3(3x+2)+11 chia hết cho 3x+2

=>11 chia hết cho 3x+2

=>3x+2 thuộc Ư(11)={11;1;-1;-11}

=>3x thuộc {9;-1;-3;-13}

=.x = {3; -0,33333 ;-1; -4,3333}   (nếu đề bài cho x thuộc Z thì x = {3;-1} )

Câu 2

       xy-5x+y=17

<=> x(y-5) +y=17

<=> x(y-5)+1(y-5)=17-5

<=>(y-5).(x+1)=12

 Vì 12=1.12= -1.(-12)=2.6= -2.(-6)=3.4= -3.(-4) nên 

Ta có bảng sau

Còn lại chắc bạn tự làm đc

Nếu ko thì mk sẽ làm tiếp cho

a) Ta có: \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)

\(\Leftrightarrow\frac{x-17-33}{33}+\frac{x-21-29}{29}+\frac{x-2\cdot25}{25}=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)

nên x-50=0

hay x=50

Vậy: x=50

b) Ta có: \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)=2\)

\(\Leftrightarrow-15x^2+46x-35+15x^2-4x-4-2=0\)

\(\Leftrightarrow42x-41=0\)

\(\Leftrightarrow42x=41\)

hay \(x=\frac{41}{42}\)

a, \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=4-4\)

\(\Leftrightarrow\left(\frac{x-17-33}{33}\right)+\left(\frac{x-21-29}{29}\right)+\left(\frac{x-2.25}{25}\right)=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\) (*)

Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\Rightarrow\) Phương trình (*) xảy ra khi: \(x-50=0\Leftrightarrow x=50\)

Vậy phương trình có nghiệm duy nhất là x = 50.

a, 1+x-10-6x=4-5x

<=> -5x-9=4-5x

<=>0x=13(vô lý)

vậy phương trình vô nghiệm

b, 6-3x+1=-3x+7

-3x+3x=7-7

<=>0x=0(luôn đúng)

vậy phương trình có vô số nghiệm

a)

\(1+x-2.\left(5+3x\right)=4-5x\)

\(\Leftrightarrow1+x-10-6x=4-5x\)

\(\Leftrightarrow x-6x+5x=4-1+10\)

\(\Leftrightarrow0x=13\)

=> Vô nghiệm

\(a,\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=5\end{matrix}\right.\\ b,\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\\ c,\Leftrightarrow\left(x+1\right)\left(3x-6\right)=0\\ \Leftrightarrow3\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow\left(x-3\right)\left(5x-10\right)=0\\ \Leftrightarrow5\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

a) \(\left(x+8\right)\left(x-5\right)=0\) \(\Rightarrow\left[{}\begin{matrix}x+8=0\\x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-8\\x=5\end{matrix}\right.\)

b) \(x\left(x-4\right)+5\left(x-4\right)=0\) \(\Rightarrow\left(x-4\right)\left(x+5\right)=0\)

     \(\Rightarrow\left[{}\begin{matrix}x-4=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

c) \(3x\left(x+1\right)-6\left(x+1\right)=0\) \(\Rightarrow\left(3x-6\right)\left(x+1\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}3x-6=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

d) \(5x\left(x-3\right)+10\left(3-x\right)=0\) \(\Rightarrow5x\left(x-3\right)-10\left(x-3\right)=0\)

     \(\Rightarrow\left(5x-10\right)\left(x-3\right)=0\)

     \(\Rightarrow\left[{}\begin{matrix}5x-10=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

a) \(=15x^3-6x^2+3x\)

b) \(=x^3-8\)

c) \(=x^2+10x+25\)

a) \(x^2-5x+6\)

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

b) \(x^2-9x+18=x^2-3x-6x+18\)

\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)

c) \(x^2-6x+5=x^2-x-5x+5\)

\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)

d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)

Câu này bạn xem lại đề nha

e) \(3x^2-5x-2=3x^2-6x+x-2\)

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

a) A= \(3x^2 - 2x+1\) với |x| = \(\dfrac{1}{2}\)

Với |x| = \(\dfrac{1}{2}\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Khi \(x=\dfrac{1}{2}\) ⇒ \(A=3.\left(\dfrac{1}{2}\right)^2-2.\dfrac{1}{2}+1=\dfrac{3}{4}-1+1=\dfrac{3}{4}\)

Khi \(x=-\dfrac{1}{2}\) ⇒ \(A=3.\left(-\dfrac{1}{2}\right)^2-2.\left(-\dfrac{1}{2}\right)+1=\dfrac{3}{4}+1+1=\dfrac{3}{4}+2=\dfrac{11}{4}\)

Vậy...

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0