Tìm x,biết
a,3(2x-1)-5(x-3)=0
b,(x-3)(x-7)-x2=1
c,5(x+3)-2x(3+x)=0
Bài 3 : Tìm x biết
a) (x-2)^2-x(x-3)=0
b) (x+3)(2x+1)-2(x-1)^2=0
c) (4x-5)^2=9(2-5x)^2
d) X^2-6x-13=0
e) (x+2)(x^2-2x+4)-x(x^2+2)=15
f) X^3-6x^2+12x-19=0
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
Tìm x,biết:
a)(2x-3)2-49=0
b)2x.(x-5)-7.(5-x)=0
c)x2-3x-10=0
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Tìm x, biết:
a) (2x-1)2+(x+3)2-5(x+7)(x-7)=0
b) x(x-5)(x+5)-(x+2)(x2-2x+4)=3
giúp tui với
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
Bài 1 giải phương trình:
a) (4x2+4x+1)-x2=0
b) x2-2x+1=4
c) x2-5x+6=0
Bài 2: giải phương trình
a) \(\dfrac{2x-5}{x+5}\)= 3
b) \(\dfrac{5}{3x+2}\)= 2x-1
c) \(\dfrac{x^2-6}{x}\)= x+\(\dfrac{3}{2}\)
d) \(\dfrac{1}{x-2}\)+3= \(\dfrac{x-3}{2-x}\)
e) \(\dfrac{3x-2}{x+7}\)=\(\dfrac{6x+1}{2x-3}\)
f) \(\dfrac{x-2}{x+2}\) - \(\dfrac{3}{x-2}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
2f.
ĐKXĐ: $x\neq \pm 2$
PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$
$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$
$\Leftrightarrow x^2-4x+4-3x-6=2x-22$
$\Leftrightarrow x^2-7x-2=2x-22$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Leftrightarrow x-4=0$ hoặc $x-5=0$
$\Leftrightarrow x=4$ hoặc $x=5$ (tm)
Tìm các số thực x, biết:
a) (2x-3)2-49=0
b) 2x(x-5)-7(5-x)=0
c) x2-3x-10=0
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a. (2x - 3)2 - 49 = 0
<=> (2x - 3)2 - 72 = 0
<=> (2x - 3 + 7)(2x - 3 - 7) = 0
<=> (2x + 4)(2x - 10) = 0
<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b. 2x(x - 5) - 7(5 - x) = 0
<=> 2x(x - 5) + 7(x - 5) = 0
<=> (2x + 7)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
c. x2 - 3x - 10 = 0
<=> x2 - 5x + 2x - 10 = 0
<=> x(x - 5) + 2(x - 5) = 0
<=> (x + 2)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a, (2x - 3)2 - 49 = 0
(2x - 3)2 - 72 = 0
(2x - 3 + 7)( 2x - 3 - 7) = 0
(2x + 4)( 2x - 10) = 0
=> 2x + 4 = 0 => 2x - 10 = 0
2x = - 4 2x = 10
x = - 2 x = 5
a)9x2 – 49 = 0
b)(x – 1)(x + 2) – x – 2 = 0
c)(4x + 1)(x - 2) - (2x -3)(2x + 1) = 7
d)x(3x + 2) + (x + 1)2 – (2x – 5)(2x + 5) = 0
e)(x + 3)(x2 – 3x + 9) –x(x – 1)(x + 1) – 27 = 0
f)(4x-3)^2-3x(3-4x)=0
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
tìm x biết
a) (2x-3)(2x+3)=0
b) x^2-1=0
c) x^2-9=0
d) 4^2-16=0
e) 25x^2-9=0
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
tìm x , biết
a) 17/6- x( x-7/6)= 7/4
b) 3/35 - ( 3/5-x)= 2/7
tìm x thuộc Z , biết
3/4-5/6 < x/12 < 1 -( 2/3-1/4)
tìm x biết
a ) 2x-3=x + 1/2
b) 4x- ( x+ 1/2) = 2x - ( 1/2 - 5 )
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
Bài 3:
a) Ta có: \(2x-3=x+\dfrac{1}{2}\)
\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)
\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)
\(\Leftrightarrow x=5\)
1. tìm x biết
a, (2x - 3)\(^2\) = |3 - 2x|
b, (x - 1)\(^2\) + (2x - 1)\(^2\) = 0
c, 5 - x\(^2\) = 1
d, x - 2\(\sqrt{x}\) = 0
g, (x - 1) + \(\dfrac{1}{7}\) = 0
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
bài 3: tìm x , biết:
a)(x+3).(2x-1)-(x-3).(x+1)=0
b)(x+4).(2x-3)-3.(x-2).(x+2)=0
c)x.(x-5).(x+5)-(x+2).(x2-2x+4)=17
a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)
\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)
b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)
\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)