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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7
<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7
<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7
<=> -8x - 4 = 7
<=> -8x = 11
<=> x = -11/8
b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8
<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8
<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8
<=> 27x + 21 = 8
<=> 27x = -13
<=> x = -13/27
c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9
<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9
<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9
<=> 3x - 6 = 9
<=> 3x = 15
<=> x = 5
d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10
<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10
<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10
<=> -11x + 6 = 10
<=> -11x = 4
<=> x = -4/11
a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)
b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)
c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)
d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(-\left(x+3\right)\left(x-4\right)+\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow-\left(x^2-4x+3x-12\right)+x^2-1=10\)
\(\Rightarrow-x^2+x+12+x^2-1=10\)
\(\Rightarrow x=10+1-12\Rightarrow x=-1\)
b, \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)
\(\Rightarrow2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)
\(\Rightarrow2x^2-5x+2-2x^2+x+21=3\)
\(\Rightarrow-4x=3-21-2\Rightarrow-4x=-20\)
\(\Rightarrow x=5\)
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Chúc bạn học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)
\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)
\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)
\(=-10x^3+19x^2+74x+1\)
\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)
\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)
\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)
\(=-5x^4-11x^3+24x^2+12x+7\)
\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)
\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)
\(=-2x^2-27x+57\)
\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)
\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)
\(=-x^3+4x^2+22x+5\)
\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)
\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)
\(=-9x^3-55x^2+4x+35\)
\(g,\left(x-1\right)^2-\left(x+2\right)^2\)
\(=x^2-2x+1-x^2-4x-4\)
\(=-6x-3\)
a) \(2x\left(x-3\right)-x\left(2x+1\right)-3\left(x+5\right)=11\)
\(\Rightarrow2x^2-6x-2x^2-x-3x-15=11\)
\(\Rightarrow-10x=26\Rightarrow x=-2,6\)
Vậy ...........
b) \(x\left(x-1\right)-\left(x^2+3x-5\right)-2\left(x+3\right)=17\)
\(\Rightarrow x^2-x-x^2-3x+5-2x-6=17\)
\(\Rightarrow-6x=18\Rightarrow x=-3\)
c) \(5x\left(x-7\right)-\left(5x+1\right)x-\left(x+3\right)2=13\)
\(\Rightarrow5x^2-35x-5x^2-x-2x-6=13\)
\(\Rightarrow-38x=19\Rightarrow x=-\frac{1}{2}\)
d) \(\left(2x^2-3x+5\right)-2x\left(x-3\right)+\left(x-1\right)\left(-2\right)=10\)
\(\Rightarrow2x^2-3x+5-2x^2+6x-2x+2=10\)
\(\Rightarrow x=3\)
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