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Answer:
\(6x^2-\left(2x+3\right)\left(3x-2\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+9x-4x-6\right)=7\)
\(\Rightarrow6x^2-\left(6x^2+5x-6\right)=7\)
\(\Rightarrow6x^2-6x^2-5x+6=7\)
\(\Rightarrow-5x+6=7\)
\(\Rightarrow-5x=1\)
\(\Rightarrow x=\frac{-1}{5}\)
\(5x\left(12+7\right)-3x\left(80x-5\right)=-100\)
\(\Rightarrow5x.19-240x^2+15x=-100\)
\(\Rightarrow95x-240x^2+15x=-100\)
\(\Rightarrow-240x^2+110x+100=0\)
\(\Rightarrow-24x^2-11x-10=0\)
\(\Rightarrow24\left(x^2-\frac{11}{24}x+\frac{121}{2304}\right)-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2-\frac{1081}{96}=0\)
\(\Rightarrow24\left(x-\frac{11}{48}\right)^2=\frac{1081}{2304}\)
\(\Rightarrow\left(x-\frac{11}{48}\right)^2=\left(\frac{\pm\sqrt{1081}}{48}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{11}{48}=\frac{\sqrt{1081}}{48}\\x-\frac{11}{48}=\frac{-\sqrt{1081}}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{1081}+11}{48}\\x=\frac{11-\sqrt{1081}}{48}\end{cases}}\)
\(\left(3x-5\right)\left(7-5x\right)-\left(5x-2\right)\left(2-3x\right)=4\)
\(\Rightarrow\left(21x-15x^2-35+25x\right)-\left(10x-15x^2-4+6x\right)-4=0\)
\(\Rightarrow36x-15x^2-35-16x+15x^2+4-4=0\)
\(\Rightarrow\left(-15x^2+15x^2\right)+\left(36x-16x\right)+\left(-35+4-4\right)=0\)
\(\Rightarrow30x-35=0\)
\(\Rightarrow x=\frac{7}{6}\)
a: \(\Leftrightarrow6x^2-6x^2+4x-9x+6=7\)
=>-5x=1
hay x=-1/5
b: \(\Leftrightarrow5x\left(12x+7\right)-3x\left(80x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-240x^2+15x=-100\)
\(\Leftrightarrow-180x^2+50x+100=0\)
hay \(x\in\left\{\dfrac{5+\sqrt{745}}{36};\dfrac{5-\sqrt{745}}{36}\right\}\)
c: \(\Leftrightarrow21x-15x^2-35+25x-\left(10x-15x^2-4+6x\right)=4\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-16x+4=4\)
=>30x-31=4
=>30x=35
hay x=7/6
a)\(5\left(2x-1\right)-4\left(8-3x\right)=7\)
\(\Leftrightarrow10x-5+12x-32=7\)
\(\Leftrightarrow22x-37=7\)
\(\Leftrightarrow22x=44\Rightarrow x=2\)
b)\(5x\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Leftrightarrow5x^2-25x-2x^2-3x=26\)
\(\Leftrightarrow3x^2-28x-26=0\)
\(\Leftrightarrow3\left(x-\dfrac{14}{3}\right)^2-\dfrac{274}{3}=0\)
\(\Rightarrow x=\dfrac{14}{3}\pm\dfrac{\sqrt{274}}{3}\)
a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)
=\(-x^5+2x^4-4x^2-1\)
f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
=\(3x^5-10x^4-13\)
b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)
=\(x^4+9x^3-11x^2+7x-2\)
f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)
=\(9x^4+5x^3-x^2-x-12\)
a )
\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)
\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)
\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)
a) \(\frac{4x + 3}{5} - \frac{6x - 2}{7}=\frac{5x + 4}{3} + 3\)
\(\Leftrightarrow \frac{21(4x +3) - 15(6x - 2)}{105}=\frac{35(5x + 4) + 315}{105}\)
<=> 21(4x + 3) - 15(6x - 2) = 35(5x + 4) + 315
<=> 84x + 63 - 90x + 30 = 175x + 140 + 315
<=> - 6x + 93 = 175x + 455
<=> - 6x - 175x = 455 - 93
<=> - 181x = 362
<=> x = - 2
Vậy ................................
b) \(\frac{3x - 11}{11} - \frac{x}{3} = \frac{3x - 5}{7} - \frac{5x - 3}{9}\)
\(\Leftrightarrow \frac{63(3x - 11) - 231x}{693}=\frac{99(3x - 5) - 77(5x - 3)}{693}\)
<=> 63(3x - 11) - 231x = 99(3x - 5) - 77(5x - 3)
<=> 189x - 693 - 231x = 297x - 495 - 385x + 231
<=> - 42x - 693 = - 88x - 264
<=> - 42x + 88x = -264 + 693
<=> 46x = 429
<=> x = \(\frac{429}{46}\)
Vậy ........
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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. |5x+4|+7=26
=>|5x+4|=19
=>5x+4=19 hoặc -19
=>5x=15
=>x=3
=>5x=-23
=>x=-23/5
b. 3-4|5-6x|=7
=>4|5-6x|=-4
=>|5-6x|=-1 (vô nghiệm)
Giá trị tuyệt đối của 1 số luôn >=0
a) |5x+4| +7 = 26
|5x+4| = 19
Th1:
5x +4 = 19
5x= 15
x=3
Th2:
5x +4 = -19
5x = -23
x= -23/5
Vậy x= 3 và x=-23/5
b) 3-4|5-6x|=7
4|5-6x|= -4
|5-6x|= -1
=> x thuộc rỗng.