Tìm x biết
5x + 3*(10 - x) + x = 45
Tìm x thuộc N , biết: 5x + 3 ( 10 - x ) + x = 45
Tìm x: 5x + 3 ( 10 - x ) + x = 45
5x+3(10-x)+x=45
<=> 5x+30-3x+x=45
<=>3x+30=45
<=>3x=15
<=>x=5
5x+3(10-x)+x=45
6x+3(10-x)=45
2x+10-x=15
x+10=15
x=15-10
x=5
5x+30-3x+x=45
5x-3x+x=15
x(5-3+1)=15
x.3=15
x=5
Tìm x: 5x + 3 ( 10 - x ) + x = 45
<=>6x+3(10-x)=45
<=>6x+30-3x=45
<=>3x+30=45
<=>3x=15
<=>x=15:3
<=>x=5
Vậy x=5
a) (x – 45).27 = 0
=> x - 45 = 0
=> x = 45
b) 23.(42- x) = 23
=> 42- x = 1
=> x = 41
c. 3x – 5=7
=> 3x = 12
=> x = 4
e. 15 – 5x=10
=> 5x = 5
=> x = 1
Tìm x: 5x + 3 ( 10 - x ) + x = 45. ( M.n giải giùm mình, mình tặng 1 tick cho! )
phương trình đơn giản vậy mà không biết giải , ngố quá
5x+30-3x+x=45
5x-3x+x+30=45
x(5-3+1)=15
x.3=15
x=5
vậy x=5
5x + 3 . ( 10 - x ) + x = 45
5x + 30 - 3x + x = 45
( 5x - 3x + x ) +30 = 45
3x = 45 - 30
3x = 15 => x = 5
BT9: Tìm x biết
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\)
\(10,\left(x+3\right)^2-x^2=45\)
\(11,\left(5x-4\right)^2-49x^2=0\)
\(12,16\left(x-1\right)^2-25=0\)
\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)
\(10,\left(x+3\right)^2-x^2=45\)
\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)
\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)
Tìm x
5x + 3 ( 10 - x ) + x = 45
GIẢI CHI TIẾT CHO TỚ NHÉ
=> 6x + 3(10 - x) = 45
=> 6x + 30 - 3x = 45
=> 3x + 30 = 45
=> 3x = 15
=> x = 5
Tìm số hữu tỉ x biết
a) (x-3)/5=6-(1/3)+[(1/3)*x]
b)[(1/2)*x]-(1/3)-5=[3-2x+7)]/4
c) [(7x)/8]-5x+45=(10/3)*x+1/4
d) [(4-x)/2007]=(7-x/2004)-(9-x/2002)-1
Tìm x,y,z biết:
Tìm x,y,z biết:
a) 7x-2y=5x-3y và 2x+3y=20
b) 2x=3y=4z-2y và x+y+z=45
c) 3x=4y-2x=7z-4y và x+y-2z=10
a.
$7x-2y=5x-3y$
$\Leftrightarrow 2x=-y$. Thay vào điều kiện số 2 ta có:
$-y+3y=20$
$2y=20$
$\Rightarrow y=10$.
$x=\frac{-y}{2}=\frac{-10}{2}=-5$
b.
$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$
$3y=4z-2y\Rightarrow 5y=4z\Rightarrow \frac{y}{4}=\frac{z}{5}$
$\Rightarrow \frac{x}{6}=\frac{y}{4}=\frac{z}{5}$
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x}{6}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{6+4+5}=\frac{45}{15}=3$
$\Rightarrow x=6.3=18; y=4.3=12; z=5.3=15$
c.
$3x=4y-2x$
$\Rightarrow 5x=4y\Rightarrow x=\frac{4}{5}y$
$3x=7z-4y$
$\Leftrightarrow \frac{12}{5}y=7z-4y$
$\Leftrightarrow \frac{32}{5}y=7z\Rightarrow z=\frac{32}{35}y$
Khi đó:
$x+y-2z=10$
$\frac{4}{5}y+y-2.\frac{32}{35}y=10$
$y.\frac{-1}{35}=10$
$y=-350$
$x=\frac{4}{5}y=\frac{4}{5}.(-350)=-280$
$z=\frac{32}{35}y=\frac{32}{35}.(-350)=-320$