tìm x biết
a) 8x-75=5x+21
b) 9x+25=-(2x-58)
c) (5-x).(x+2)=0
d) 3./x+3/ -4=-7-(-18)
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a, \(\left|x-3\right|-16=-4\Leftrightarrow\left|x-3\right|=12\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}}\)
b, \(26-\left|x+9\right|=-13\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=39\\x+9=-39\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30\\x=-48\end{cases}}}\)
f, \(9x+25=-\left(2x-58\right)\Leftrightarrow9x+25=-2x+58\)
\(\Leftrightarrow11x-33=0\Leftrightarrow x=3\)
Làm nốt nhé !
a. \(\left|x-3\right|-16=-4\)
\(\Rightarrow\left|x-3\right|=12\)
\(\Rightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}\)
b. \(26-\left|x+9\right|=-13\)
\(\Rightarrow\left|x+9\right|=39\)
\(\Rightarrow\orbr{\begin{cases}x+9=39\\x+9=-39\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=30\\x=-48\end{cases}}\)
c. \(5-\left(10-x\right)=7\)
\(\Rightarrow10-x=-2\)
\(\Rightarrow x=12\)
d. \(11+\left(15-x\right)=1\)
\(\Rightarrow15-x=-10\)
\(\Rightarrow x=25\)
e. \(8x-75=5x+21\)
\(\Rightarrow8x-5x=75+21\)
\(\Rightarrow3x=96\)
\(\Rightarrow x=32\)
f. \(9x+25=-\left(2x-58\right)\)
\(\Rightarrow9x+25=-2x+58\)
\(\Rightarrow9x+2x=58-25\)
\(\Rightarrow11x=33\)
\(\Rightarrow x=3\)
g. \(\left|x+3\right|=4\)
\(\Rightarrow\orbr{\begin{cases}x+3=4\\x+3=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=-7\end{cases}}\)
h. \(15-\left|2x-1\right|=\left|-8\right|\)
\(\Rightarrow15-\left|2x-1\right|=8\)
\(\Rightarrow\left|2x-1\right|=7\)
\(\Rightarrow\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
a) |2x-1|=|-7|
<=>|2x-1|=7
<=>\(\left[\begin{matrix}2x-1=7\\-\left(2x-1\right)=7\end{matrix}\right.\)
<=>\(\left[\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy: x=-3 hoặc x=4
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
a) 8x-75=5x+21
8x-5x=21+75
3x=96:3
x=32
b) 9x+25=-(2x-58)
9x+25=-2x+58
9x+2x=58-25
11x=33
x=33:11
x=3
tick mình nha
Mình không viết lại đề nhé!
a) 27 - x + 15 + x = x + 2
-x + x - x = 2 - 15 - 27
-x = -40
x = 40
b) 8x - 75 = 5x +21
8x - 5x = 21 + 75
3x = 96
x = 32
c) 15 - |2x - 1| = |-8|
15 - 2x + 1 = 8
-2x = 8 -1 -15
-2x = -8
x = 4
d) 9x + 25 = -(2x - 58)
9x + 25 = -2x + 58
9x +2x = 58 - 25
7x = 33
x = 33/7
Lời giải:
a. Đề có cả x,y. Bạn xem lại
b.
PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$
$\Leftrightarrow (x-3)(5x-2)=0$
$\Leftrightarrow x-3=0$ hoặc $5x-2=0$
$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$
c.
PT $\Leftrightarrow (7x-2)(x-4)=0$
$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$
$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$
d. Đề thiếu.
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12
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
\(a,8x-75=5x+21\)
\(8x-5x=21+75\)
\(3x=96\)
\(x=32\)
\(b,9x+25=-\left(2x-58\right)\)
\(9x+25=-2x+58\)
\(9x+2x=58-25\)
\(11x=33\)
\(x=3\)
\(c,\left(5-x\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}}\)
a) \(8x-75=5x+21\)
\(8x-5x=21+75\)
\(3x=96\)
\(x=32\)
vậy \(x=32\)
b) \(9x+25=-\left(2x-58\right)\)
\(9x+25=-2x+58\)
\(9x+2x=58-25\)
\(11x=33\)
\(x=3\)
vậy \(x=3\)