Tìm x
a) 2 . ( x - 1 ) + 3 ( x - 2 ) = x - 4
b) 3 . ( 4 - x ) - 2 . ( x - 1 ) = x + 20
c) 135 - | 9 - x | = 35
d) | 2x + 3 | = 5
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a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x-5\right)=-4\)
\(\Leftrightarrow x^2+5x+6-x^2+7x-10=-4\)
\(\Leftrightarrow12x=0\)
hay x=0
b: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow x^3+1-x^3+9x=8\)
\(\Leftrightarrow9x=7\)
hay \(x=\dfrac{7}{9}\)
c: Ta có: \(4x^2-9=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x+1-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
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 \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\)
hay \(x=-\dfrac{1}{2}\)
c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)
hay x=0
a) (x+3)^2 - (2-x)^2 = 1
x^2 + 6x + 9 - (4 - 4x + x^2) = 1
x^2 + 6x + 9 - 4 + 4x - x^2 = 1
10x + 5 = 1
10x = -4
x = -4/10
x = -2/5
Vậy giá trị của x là -2/5.
b) 5(x-2)^2 - (x+3)^2 = (2x-1)^2
5(x^2 - 4x + 4) - (x^2 + 6x + 9) = 4x^2 - 4x + 1
5x^2 - 20x + 20 - x^2 - 6x - 9 = 4x^2 - 4x + 1
4x^2 - 26x + 30 = 4x^2 - 4x + 1
-26x + 30 = -4x + 1
-22x = -29
x = 29/22
Vậy giá trị của x là 29/22.
c) (x-1)^2 - x(x+5)^2 = 7
x^2 - 2x + 1 - x(x^2 + 10x + 25) = 7
x^2 - 2x + 1 - x^3 - 10x^2 - 25x = 7
-x^3 - 9x^2 - 27x - 6 = 0
d) (3x-2)^2 - 9(x+2)^2 = 3
9x^2 - 12x + 4 - 9x^2 - 36x - 36 = 3
-48x - 32 = 3
-48x = 35
x = -35/48
Vậy giá trị của x là -35/48.
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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c) 135 - | 9 - x | = 3
| 9 - x | = 135 - 3
| 9 - x | = 132
=> 9 - x = 132
x = 9 - 132
x = 123
hoặc 9 - x = - 132
x = 9 - ( - 132 )
x = 9 + 132
x = 141
Vậy ..................
c) 135 - | 9 - x | = 35
\(2\left(x-1\right)+3\left(x-2\right)=x-4\)
\(2x-2+3x-6=x-4\)
\(5x-8=x-4\)
\(5x-8-x+4=0\)
\(4x-4=0\)
\(4x=4\)
\(x=1\)
\(3\left(4-x\right)-2\left(x-1\right)=x+20\)
\(12-x-2x+2=x+20\)
\(14-3x=x+20\)
\(14-3x-x-20=0\)
\(-6-4x=0\)
\(4x=-6\)
\(x=-\frac{3}{2}\)
\(\left|2x+3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}2x=2\\2x=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}}\)