tim x biet
(x-2)^2=4x^2+4x+1
(tim x biet (x+1)(2-x)-(3x+5)(x+2)=-4x^2+1
Tim x biet
|x+1|+|x+2|+|x+3|=4x
Nguyễn Thị Quỳnh Linh
Ta có :
\(\left|x+1\right|\ge0\)
\(\left|x+2\right|\ge0\)
\(\left|x+3\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+3\right|\ge0\)
\(\Rightarrow4x\le0\)
Mà 4 > 0
=> x > 0
=> x + 1 + x + 2 + x + 3 = 4x (phá trị uyệt đối vì x dương)
=> 3x + 6 = 4x
=> 4x - 3x = 6
=> x = 6
\(\left|x+1\right|+\left|x+2\right|+\left|x+3\right|=4x\)
\(3x+6=4x\Rightarrow x=6\)
| x + 1 | \(\ge\)0 ; | x + 2 | \(\ge\)0 ; | x + 3 | \(\ge\)0
=> | x + 1 | + | x + 2 | + | x + 3 | \(\ge\)0
=> 4x \(\ge\)=> x \(\ge\)0
=> x+1 + x+2 + x +3= 4x
=> 3x + 6 = 4x
=> x = 6
Vậy....
tim x biet:
( x^2-4x+16 )( x+4 )-x ( x+1 )(x+2)+3x^2=0
(8x+2)(1-3x)+(6x-1)(4x-10)=-50
( x2 - 4x + 16 )( x + 4 ) - x( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 43 - x( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = -50
<=> 2x - 24x2 + 2 + 24x2 - 64x + 10 = -50
<=> -62x + 12 = -50
<=> -62x = -62
<=> x = 1
tim x biet :
a) ( 2x^2+4 ) - (x^2-3/2 ) = (-3+4x^2)+ ( -4x^2/3+1)
b) ( I xI -4) - ( 2- I-xI ) = 1/3 . I xI -5
Tim x biet(4x-3)-(x+5)=(x+2)-2×(x-10)
mình không biết cách làm, nhưng mình biết x = 7,5
\(4x-3-x-5-=x+2-2x+2.10\)
\(3x-8=x+2-2x+20\)
\(3x=x+10-2x+20\)
\(3x=-x+30\)
\(4x=30\)
\(x=\frac{30}{4}=7,5\)
Cam on cac b mk cx ra ket qua la 7,5
tim x; y biet 4x=5y va \(x^2-y^2=1\)
Vì \(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)
\(\Rightarrow\left\{\begin{matrix}\frac{x}{5}=\frac{1}{9}\\\frac{y}{4}=\frac{1}{9}\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=\frac{5}{9}\\y=\frac{4}{9}\end{matrix}\right.\)
Từ 4x = 5y
=> \(\frac{x}{5}=\frac{y}{4}\) ( từ đẳng thức suy ra tỉ lệ thức )
\(=>\frac{x^2}{5^2}=\frac{y^2}{4^2}=>\frac{x^2}{25}=\frac{y^2}{16}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{25}=\frac{y^2}{16}=\frac{x^2-y^2}{25-16}=\frac{1}{9}\)
Do đó:
\(\frac{x}{5}=\frac{1}{9}=>x=5:9=\frac{5}{9}\)
\(\frac{y}{4}=\frac{1}{9}=>y=4:9=\frac{4}{9}\)
Vậy x = \(\frac{5}{9}\) và y = \(\frac{4}{9}\)
Giải:
Ta có: \(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\)
Đặt \(\frac{x}{5}=\frac{y}{4}=k\)
\(\Rightarrow x=5k,y=4k\)
Mà \(x^2-y^2=1\)
\(\Rightarrow\left(5k\right)^2-\left(4k\right)^2=1\)
\(\Rightarrow5^2.k^2-4^2.k^2=1\)
\(\Rightarrow k^2\left(5^2-4^2\right)=1\)
\(\Rightarrow k^2.9=1\)
\(\Rightarrow k^2=\frac{1}{9}\)
\(\Rightarrow k=\pm\frac{1}{3}\)
+) \(k=\frac{1}{3}\Rightarrow x=\frac{5}{3};y=\frac{4}{3}\)
+) \(k=\frac{-1}{3}\Rightarrow x=\frac{-5}{3};y=\frac{-4}{3}\)
Vậy cặp số \(\left(x;y\right)\) là \(\left(\frac{5}{3};\frac{4}{3}\right);\left(\frac{-5}{3};\frac{-4}{3}\right)\)
tim x biet : 2|4x-2| - 2x = 14
\(\left[{}\begin{matrix}2\left(4x-2\right)-2x=14\left(x\ge\dfrac{1}{2}\right)\\-2\left(4x-2\right)-2x=14\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=18\left(x\ge\dfrac{1}{2}\right)\\-10x=10\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(x\ge\dfrac{1}{2}\right)\left(TM\right)\\x=-1\left(x< \dfrac{1}{2}\right)\left(TM\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
tim x biet :
a) ( 2x2+4 ) - (x2-3/2 ) = (-3+4x2)+ ( -4x2/3+1)
b) ( I xI -4) - ( 2- I-xI ) = 1/3 . I xI -5
Không có ngiệm nguyên hay hữu tỉ mà bạn
tim x biet :
a) ( 2x2+4 ) - (x2-3/2 ) = (-3+4x2)+ ( -4x2/3+1)
b) ( I xI -4) - ( 2- I-xI ) = 1/3 . I xI -5