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+) (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
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(\Rightarrow-80x=-480\Rightarrow x=6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)
\(\Rightarrow-4x=0\Rightarrow x=0\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow10x-12x-12x=-16+11+16-15\)
\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)
\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)
\(\Rightarrow5x-12x+24x-90x+36=182\)
\(\Rightarrow-73x=182-36\)
\(\Rightarrow-73x=146\Rightarrow x=-2\)
Chúc bạn học tốt!!!
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
f) |-2x| = 3x+4
⇒\(\left[{}\begin{matrix}-2x=3x+4\\-2x=-3x-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-4}{5}\\x=-4\end{matrix}\right.\)
g) |2x-1| = 6-x
⇒\(\left[{}\begin{matrix}2x-1=6-x\\2x-1=x-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-5\end{matrix}\right.\)
h) |-1+5x| = 8-x
⇒\(\left[{}\begin{matrix}-1+5x=8-x\\-1+5x=x-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-7}{4}\end{matrix}\right.\)
i) |-2x+1| = x+3
⇒\(\left[{}\begin{matrix}-2x+1=x+3\\-2x+1=-x-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=4\end{matrix}\right.\)
k) |-2-5x| = -4x+7
⇒\(\left[{}\begin{matrix}-2-5x=-4x+7\\-2-5x=4x-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-9\\x=\frac{5}{9}\end{matrix}\right.\)
a) |x-2| = 3
⇒\(\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b) |x+1| = |2x+3|
⇒\(\left[{}\begin{matrix}x+1=2x+3\\x+1=-2x-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\frac{-4}{3}\end{matrix}\right.\)
c) |3x| = x+6
⇒\(\left[{}\begin{matrix}3x=x+6\\3x=-x-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3}{2}\end{matrix}\right.\)
d) |x-5| = 13-2x
⇒\(\left[{}\begin{matrix}x-5=13-2x\\x-5=2x-13\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=8\end{matrix}\right.\)
e) |5x-1| = x-12
⇒\(\left[{}\begin{matrix}5x-1=x-12\\5x-1=12-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-11}{4}\\x=\frac{13}{6}\end{matrix}\right.\)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Leftrightarrow156-56x=24x-324\)
\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)
\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)
\(\Leftrightarrow7x+37=11x-35\)
\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)
\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)
\(\Leftrightarrow5x+33x-18-182=0\)
\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)
Bạn xét các Trường hợp biểu thức trong dấu GTTĐ lớn hơn hoặc 0 và bé hơn 0 sau đó giải ra trừ cái đầu
A,\(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
B,Xét x+1\(\ge\)0\(\Leftrightarrow x\ge-1\)
khi dó \(|x+1|=x-12\)
\(\Leftrightarrow\) \(x+1=x-12\)
\(\Leftrightarrow0x=-13\left(PT.vô.nghiệm\right)\)
xét x+1<0\(\Leftrightarrow x< -1\)
ki dó\(|x+1|=x-12\)
\(\Leftrightarrow-x-1=x-12\)
\(\Leftrightarrow-2x=-11\)
\(\Leftrightarrow x=\frac{11}{2}\)(loại)
vậy ko có giá trị x t/m ĐKBT
câu 6: \(|-2|=3x+4\)
\(\Leftrightarrow2=3x+4\)
\(\Leftrightarrow3x=-2\)
\(\Leftrightarrow x=-\frac{2}{3}\)
câu cuối
\(|-2-5x|=-4x+7\)
xét -2-5x\(\ge0\Leftrightarrow x\ge-\frac{2}{5}\)
Khi dó \(|-2-5x|=-4x+7\)
\(\Leftrightarrow-2-5x=-4x+7\)
\(\Leftrightarrow x=-9\)(loại)
Xét -2+5x<0\(\Leftrightarrow x< \frac{2}{5}\)
\(\Leftrightarrow2+5x=-4x+7\)
\(\Leftrightarrow x=\frac{5}{9}\)(loại)
VÂỵ không có gia trị x t/m đk
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