1. Tìm x biết
a/ (2x+3)(x-4) + (x-5)(x-2)=(3x-5)(x-4)
b/ (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)
Các bn nhớ giải rõ ràng nha, ai giải tốt nhất mk sẽ tick cho
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a)\(2x^2\)+\(3\left(x^2-1\right)\)=\(5x\left(x+1\right)\)
\(2x^2\)+\(3x^2\)\(-3\)=\(5x^2+5x\)
\(5x^2-5x^2-5x=3\)
\(-5x=3\)
\(x=\frac{-3}{5}\)
tự ghi dấu suy ra ở đằng trước nhé
b) Vì \(2x\left(5-3x\right)=2x\left(3x-5\right)-3\left(x-7\right)=3\)
nên chỉ cần giải: \(6x^2-10x-3x+21=3\)
\(\Leftrightarrow6x^2-13x+21=3\)
\(\Leftrightarrow6x^2-13x+18=0\)
\(\Rightarrow\)pt vô nghiệm
a, Xét : x-4 = 0 => x= 4
2x+1 = 0 => x= \(\frac{1}{2}\)
x+3 = 0 => x = -3
x + 9 = 0 => x = -9
Khi đó ta có bảng xét dấu :
x | -9 | -3 | \(\frac{1}{2}\) | 4 |
x-4 | -13 | -7 | \(\frac{-7}{2}\) | 0 |
2x+1 | -17 | -5 | 2 | 9 |
x+3 | -6 | 0 | \(\frac{7}{2}\) | 7 |
x+9 | 0 | 6 | \(\frac{19}{2}\) | 13 |
=> có 5 trường hợp:
TH1 : \(x\le-9\)
TH2 : \(-9\le x< -3\)
TH3 : \(-3\le x< \frac{1}{2}\)
TH4 : \(\frac{1}{2}\le x< 4\)
Do đó :
TH1 : \(x\le-9\)
Ta có : /x-4/ = -(x-4) = 4 - x
/2x+1/ = -(2x+1) = -2x -1
/x+3/ = -(x + 3 ) = -x - 3
/x-9/ = -(x-9) = -x + 9 Thay vào đề bài ta có:
3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5
=> 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5
=>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5
=> -13 - 7x = 5
7x = -13 - 5
7x = -18
x = \(\frac{-18}{7}\)( Ko TM)
Tương tự với 4 trường hợp còn lại.
a: \(x^3+8x=5x^2+4\)
=>\(x^3-5x^2+8x-4=0\)
=>\(x^3-x^2-4x^2+4x+4x-4=0\)
=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2=0\)
=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: \(x^3+3x^2=x+6\)
=>\(x^3+3x^2-x-6=0\)
=>\(x^3+2x^2+x^2+2x-3x-6=0\)
=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)
3: ĐKXĐ: x>=0
\(2x+3\sqrt{x}=1\)
=>\(2x+3\sqrt{x}-1=0\)
=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)
=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)
=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)
=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)
=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)
4: \(x^4+4x^2+1=3x^3+3x\)
=>\(x^4-3x^3+4x^2-3x+1=0\)
=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)
=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)
=>(x-1)^2=0
=>x-1=0
=>x=1
a.
\(x^3+8x=5x^2+4\)
\(\Leftrightarrow x^3-5x^2+8x-4=0\)
\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
b.
\(x^3+3x^2-x-6=0\)
\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)
\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)
(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)
20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1
10x2−19x−33=010x2−19x−33=0
(10x+11)(x−3)=0
chỉ bt lm con b thoy
..army,,,,,,,,,,
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)
\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\frac{22}{5}\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)
\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)
\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)
\(\Leftrightarrow20x^2-16x-33=10x^2\)
\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)
\(\Leftrightarrow20x^2-16x-33=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)