tìm x :
1) (x-5)2 - (x+3)(x-3) =14
2) (x+7)2 - 3x - 21 = 0
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a) | 5/4x -7/2| - | 5/8x + 3/5| = 0
|5/4x - 7/2| = | 5/8x + 3/5|
TH1: 5/4x - 7/2 = 5/8x + 3/5
=> 5/4x - 5/8x = 3/5 +7/2
5/8x = 41/10
x = 41/10:5/8
x = 164/25
TH2: 5/4x - 7/2 = -5/8x - 3/5
=> 5/4x + 5/8x = -3/5 +7/2
15/8x = 29/10
x = 29/10 : 15/8
x = 116/75
KL: x = 164/25 hoặc x = 116/75
các bài cn lại b lm tương tự nha! h lm dài lắm!
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
1) 2x(x + 1) - x2(x + 2) + x3 - x + 4 = 0
<=> 2x.x + 2x.1 + (-x2).x + (-x2).2 + x3 - x + 4 = 0
<=> 2x2 + 2x - x3 - 2x2 + x3 - x = 0 - 4
<=> x = -4
=> x = -4
2) xem lại đề rồi chúng mình nói chuyện cậu nha :))
3) tương tự (mình hơi lười, thông cảm :v)
3, [(3x - 5)(7 - 5x)] - [(5x + 2)(2 - 3x)] = 4
<=> ( 21x -15x^2 -35 +25x) - (10x -15x^2 + 4-6x)=4
<=> 21x -15x^2 -35 +25x- 10x + 15x^2 - 4+6x =4
<=> 42x - 39 =4
<=> 42x = 43
<=< x =43/42
2, (3x - 2)(4x - 5 ) - (2x - 1)(6x + 2) = 0
12x2- 15x - 8x + 10 - 12x2 - 4x + 6x + 2 = 0
- 21x = -12
x = 4/7
1, đã có người giải
d) |2x – 5| –7 = 22
| 2x -5 | = 22+7
| 2x -5 | = 29
TH1: 2x-5 = -29
2x = -29+5
2x= -24
x= -24:2
x= -12
TH2: 2x -5 =29
2x = 29+5
2x= 34
x= 34:2
x= 17
Vậy...
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x2 - 16x - 34 = 10x2 + 3x - 34
=> 10x2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0
hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10
Vậy x = 0 ; x = 19/10
Rút gọn hết ta được :
a/ 41x - 17 = -21
=> 41x = -4 => x = 4/41
b/ 34x - 17 = 0
=> 34x = 17
=> x = 17/34 = 1/2
c/ 19x + 56 = 52
=> 19x = -4
=> x = -4/19
d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34
=> 10x 2 - 19x = 0
=> x(10x - 19) = 0
=> x = 0 hoặc 10x - 19 = 0
=> 10x = 19
=> x = 19/10
Vậy x = 0 ; x = 19/10
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
1) (x - 5)2 - (x + 3)(x - 3) = 14
=> x2 - 10x + 25 - x2 + 9 = 14
=> -10x + 34 = 14
=> -10x = 14 - 34
=> -10x = -20
=> x = 2
2) (x + 7)2 - 3x - 21 = 0
=> x2 + 14x + 49 - 3x - 21 = 0
=> x2 + 11x + 28 = 0
=> x2 + 4x + 7x + 28 = 0
=> x(x + 4) + 7(x + 4) = 0
=> (x + 7)(x + 4) = 0
=> \(\orbr{\begin{cases}x+7=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-7\\x=-4\end{cases}}\)
a
\(\left(x-5\right)^2-\left(x+3\right)\left(x-3\right)=14\)
\(\Leftrightarrow x^2-10x+25-x^2+9=14\)
\(\Leftrightarrow-10x=-20\)
\(\Leftrightarrow x=2\)
b
\(\left(x+7\right)^2-3x-21=0\)
\(\Leftrightarrow x^2+14x+49-3x-21=0\)
\(\Leftrightarrow x^2+11x+28=0\)
\(\Leftrightarrow\left(x^2+4x\right)+\left(7x+28\right)=0\)
\(\Leftrightarrow x\left(x+4\right)+7\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=0\)
\(\Leftrightarrow x=-4;x=-7\)