20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)

⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0

⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0

+   x−1=0⇔x=1x−1=0⇔x=1

+    2x+11=0⇔x=−5,52x+11=0⇔x=−5,5

Phương trình có nghiệm x = 1 hoặc x = -5,5

b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0

⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0

⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0

+     15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)

+      5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)

Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

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

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

Đặt\(3x^2+5x-8=a\)\(\Rightarrow3x^2+5x+1=a+9\), từ đó phương trình có dạng \(\sqrt{a}-\sqrt{a+9}=1\), bn giải phương trình bình thường tìm ra a, rồi suy ra x nha!. Học tốt

Mấy cái này chuyển vế đổi dấu là xong í mà :3

1,

16-8x=0

=>16=8x

=>x=16/8=2

2, 

7x+14=0

=>7x=-14

=>x=-2

3,

5-2x=0

=>5=2x

=>x=5/2

Mk làm 3 cau làm mẫu thôi

Lúc đăng đừng đăng như v :>

chi ra khỏi ngt nản

từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại

1, 16 - 8x = 0

<=>-8x = 16

<=> x = -2

Vậy_

2, 7x + 14 = 0

<=> 7x = -14

<=> x = -2

3, 5 - 2x = 0

<=> - 2x = -5

<=> x =\(\frac{5}{2}\)

Vậy_

4, 3x - 5 = 7

<=> 3x = 7 + 5

<=> 3x = 12

<=> x = 4

Vậy...

5, 8 - 3x = 6

<=> - 3x = 6 - 8

<=> -3x = - 2

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

Vậy......

b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

 Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)

  Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)

a)(x-1)(5x+3)=(3x-8)(x-1)

\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0

\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)

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

\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)

\(a.\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8.\)

\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)

\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)

\(\Leftrightarrow-4x+3=8\)

\(\Leftrightarrow-4x=5\)

\(\Leftrightarrow x=-\frac{5}{4}\)

\(b.\left(3x-5\right)\left(7-5x\right)+\left(5x-2\right)\left(3x-2\right)-2=0\)

\(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)

\(\Leftrightarrow42x-41=0\)

\(\Leftrightarrow42x=41\)

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

a, (10x+9) x -(5x-1) (2x +3 )=8

= 10x²+ 9x-(10x² +15x -2x -3) =8

= 10x²+9x - 10x^{2 -}13x +3 =8

= - 4 x +3 =8

= - 4 x =5

suy ra x= -5/4

b, (3x -5)(7-5x)+(5x+2) (3x -2) -2 =0

= 21x -15x² - 35 +25x + 15x² -10x +6x -4 -2 =0

=42x -41 =0

suy ra x= 41/42

\(a.6x-3=5x+2\) 

\(\Leftrightarrow6x-3-5-2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

\(S=\left\{1\right\}\)

\(b.2-3x=5x-6\)

\(\Leftrightarrow2-3x-5x+6=0\)

\(\Leftrightarrow-8x+8=0\) 

\(\Leftrightarrow x=1\)

\(S=\left\{1\right\}\)

\(c.\left|3x\right|=2x+7\left(1\right)\)

\(TH_1:3x\ge0\Leftrightarrow x\ge0\)

\(\left(1\right)\Leftrightarrow3x=2x+7\)

      \(\Leftrightarrow3x-2x=7\) 

      \(\Leftrightarrow x=7\left(n\right)\) 

\(TH_2:3x< 0\Leftrightarrow x< 0\)

\(\left(1\right)\Leftrightarrow-3x=2x+7\)

      \(\Leftrightarrow-3x-2x=7\) 

      \(\Leftrightarrow-5x=7\) 

       \(\Leftrightarrow x=\dfrac{-5}{7}\left(n\right)\) 

Vậy pt (1) có tập n₀ S = \(\left\{7,\dfrac{-5}{7}\right\}\)

\(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)

Đặt: \(\left\{{}\begin{matrix}a=\sqrt{3x^2+5x+8}\\b=\sqrt{3x^2+5x+1}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2=3x^2+5x+8\\b^2=3x^2+5x+1\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}a-b=1\\a^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\\left(1+b\right)^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\1+2b+b^2-b^2=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1+b\\b=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3x^2+5x+8}=4\\\sqrt{3x^2+5x+1}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+5x+8=16\\3x^2+5x+1=9\end{matrix}\right.\) \(\Leftrightarrow3x^2+5x-8=16\) \(\Leftrightarrow3x^2-3x+8x-8=0\)

\(\Leftrightarrow3x\left(x-1\right)+8\left(x-1\right)=0\) \(\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(n\right)\\x=-\dfrac{8}{3}\left(n\right)\end{matrix}\right.\)

\(\sqrt{3x^2+5x+8}-\sqrt{3x^2+6x+1}=1\)

Đặt : \(3x^2+5x+8=a\) . Phương trình trở thành :

\(\sqrt{a}-\sqrt{a-7}=1\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{a-7}\right)^2=1\)

\(\Leftrightarrow a-2\sqrt{a\left(a-7\right)}+a-7=1\)

\(\Leftrightarrow2a-2\sqrt{a\left(a-7\right)}=8\)

\(\Leftrightarrow2\sqrt{a\left(a-7\right)}=2a-8\)

\(\Leftrightarrow\sqrt{a\left(a-7\right)}=a-4\)

\(\Leftrightarrow a\left(a-7\right)=\left(a-4\right)^2\)

\(\Leftrightarrow a^2-7a=a^2-8a+16\)

\(\Leftrightarrow a=16\)

\(\Leftrightarrow3x^2+5x+8=16\)

\(\Leftrightarrow3x^2+5x-8=0\)

\(\Delta=5^2+4.3.8=25+96=121>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-5+\sqrt{121}}{6}=1\\x_2=\dfrac{-5-\sqrt{121}}{6}=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{8}{3};1\right\}\)