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20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(\Rightarrow-80x=-480\Rightarrow x=6\)
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!!!
1. 3y = 0
=> y = 0
2. 1+x = 0
<+ x = -1
3.
\(1-2t=0\)
\(\Leftrightarrow2t=1\)
\(\Leftrightarrow\dfrac{1}{2}\)
4. 2x +x + 3 =0
\(\Leftrightarrow3x+3=0\)
\(\Leftrightarrow x=-3\)
5.
\(25x-20=0\)
\(\Leftrightarrow25x=20\)
\(\Leftrightarrow x=\dfrac{4}{5}\)
7.
2x-3 = x+5
<=> 2x - x = 5+3
<=> x = 8
8.
x-8=2x+3
<=> x - 2x = 3+8
<=> -x = 11
<=> x = -11
9. 17-2x = 3x-5
<=> -2x-3x = -5-17
<=> -5x = -22
<=> x = \(\dfrac{22}{5}\)
10.
2x+x+22=0
<=> 3x+22=0
<=> 3x = -22
<=> x = \(\dfrac{-22}{3}\)
Mấy bài kia tự giải tương tự nhá!!!
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, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)
\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)
\(\Leftrightarrow5x+33x-18-182=0\)
\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)
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
5/ (x2 – 4) + (x – 2)(4 – 2x) = 0
⇔(x-2)(x+2)+(x – 2)(4 – 2x)=0
⇔(x-2)(x+2+4-2x)=0
⇔(x-2)(6-x)=0
⇔\(\left[{}\begin{matrix}x-2=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
6/ x(2x – 7) – 4x + 14 = 0
⇔2x2-11x+14=0
⇔(x-\(\frac{7}{2}\))(x-2)=0
⇔\(\left[{}\begin{matrix}x-\frac{7}{2}=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
7/ x2 – x – (3x–3)= 0
⇔x2-4x+3=0
⇔(x-3)(x-1)=0
⇔\(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
8/ (x2 – 2x + 1) – 4 = 0
⇔(x-1)2-4=0
⇔(x-1-4)(x-1+4)=0
⇔(x-5)(x+3)=0
⇔\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
9/ 4x2 + 4x + 1 = x2
⇔3x2+4x+1=0
⇔(3x+1)(x+1)=0
⇔\(\left[{}\begin{matrix}3x+1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-1\end{matrix}\right.\)
10/ x2 – x = - 2x + 2
⇔3x2-x-2=0 (chuyển vế)
⇔(3x+2)(x-1)=0
⇔\(\left[{}\begin{matrix}3x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=1\end{matrix}\right.\)
11/ x2 – 5x + 6 = 0
⇔x2-3x-2x+6=0
⇔x(x-3)-2(x-3)=0
⇔(x-3)(x-2)=0
⇔\(\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Mình làm bài khá tắt nên có gì không hiểu bạn cứ hỏi mình nha!
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Theo bài ra, ta có:
\(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\\ \left(4,5-2x\right).\dfrac{11}{7}=\dfrac{11}{14}\\ 4,5-2x=\dfrac{11}{14}:\dfrac{11}{7}\\ 4,5-2x=\dfrac{1}{2}\\ 2x=4,5-\dfrac{1}{2}\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy x = 2
\(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Leftrightarrow\left(4,5-2x\right).\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\dfrac{11}{7}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}.\dfrac{7}{11}\)
\(\Leftrightarrow4,5-2x=\dfrac{1}{2}\)
\(\Leftrightarrow-2x=\dfrac{1}{2}-4,5\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy phương trình có ngiệm x = 2 .