Tìm x biết 7x^2-7x(x-3)=x+2
Làm nhanh hộ mink nhé hihi!!!!!!!!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
tìm x,y biết 7x=4y và y-x =24 giải hộ mink nha làm theo tính chất dãy tỉ số = nhau lời giải chi tiết
7x = 4y nên \(\frac{y}{\frac{1}{4}}=\frac{x}{\frac{1}{7}}=\frac{y-x}{\frac{1}{4}-\frac{1}{7}}=\frac{24}{\frac{3}{28}}=224\)=> x = 224 : 7 = 32 ; y = 224 : 4 = 56
7x=4y => y/7 = x/4
ap dung day ty so = nhau ta co;
(y-x) /(7-4) = 24/3 =8
x= 4.8 = 32
y = 7.8 = 56
Ta có: |x+1| ;|x+3| ;|x+5|>=0
=> |x+1|+|x+3|+|x+5|>=0
=> 7x>=0
=> x+1+x+3+x+5=7x
3x+8=7x
4x=8
x=2
|x + 1| + |x + 3| + |x + 5| = 7x
có |x + 1| > 0; |x + 3| > 0; |x + 5| > 0
=> 7x > 0
=> x > 0
=> x + 1 + x + 3 + x + 5 = 7x
=> 3x + 9 = 7x
=> 3x - 7x = - 9
=> -4x = -9
=> x = 9/4
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
\(\text{a. }-15+x=-4\)
\(\Rightarrow x=-4+15\)
\(\Rightarrow x=11\)
\(\text{Vậy }x=11\)
\(\text{b.}35-x=-12-3\)
\(\Rightarrow35-x=-15\)
\(\Rightarrow x=35+15\)
\(\Rightarrow x=50\)
\(\text{Vậy }x=50\)
\(\text{c.}11x-7x+x=325\)
\(\Rightarrow5x=325\)
\(\Rightarrow x=325\div5\)
\(\Rightarrow x=65\)
\(\text{Vậy }x=65\)
\(\text{d.}15-\left|x-3\right|=7\)
\(\Rightarrow\left|x-3\right|=15-7\)
\(\Rightarrow\left|x-3\right|=8\)
\(\Rightarrow\orbr{\begin{cases}x-3=-8\\x-3=8\end{cases}}\)
\(\text{Trường hợp : }x-3=-8\)
\(\Rightarrow x=-8+3\)
\(\Rightarrow x=-5\)
\(\text{Trường hợp : }x-3=8\)
\(\Rightarrow x=8+3\)
\(\Rightarrow x=11\)
\(\text{Vậy }x\in\left\{-5;11\right\}\)
a,-15+x=-4 b,35-x=-12-3
x=-4+15 35-x=-12+(-3)
x=11 35-x=-15
vậy x=11 x=35+15
x=50
vẫy=50
a.
-2 (x+6)+6 (x-10)=8
-2x -12 +6x -60 =8
4x -72 =8
4x = -64
x= -16
b.
7x (2+x) -7x (x+3) =14
7x [ (2+x)-(x+3) ] =14
7x (2+x-x-3) =14
-7x =14
x= -2
\(1,\Rightarrow3^{x-3}=\left(3^2\right)^8:\left(3^3\right)^5=3^{16}:3^{15}=3^1\\ \Rightarrow x-3=1\\ \Rightarrow x=4\\ 2,\Rightarrow7^x\left(1+7^2\right)=350\\ \Rightarrow7^x=\dfrac{350}{50}=7=7^1\\ \Rightarrow x=1\)
\(3,\Rightarrow2^{2+2x+2}-2^{2x}=240\\ \Rightarrow2^{2x}\left(2^4-1\right)=240\\ \Rightarrow2^{2x}=\dfrac{240}{15}=16=2^4\\ \Rightarrow2x=4\Rightarrow x=2\)
Bài 1:
1) \(\left|x-15\right|+x-15=0\)\(\Leftrightarrow\)\(\left|x-15\right|=15-x\)
+ Với \(x\ge15\forall x\)\(\Leftrightarrow\)\(x-15\ge0\forall x\)\(\Rightarrow\)\(\left|x-15\right|=x-15\)
\(\Rightarrow x-15=15-x\)
\(\Leftrightarrow2x=30\)
\(\Leftrightarrow x=15\)( thỏa mãn điều kiện )
+ Với \(x< 15\forall x\)\(\Leftrightarrow\)\(x-15< 0\forall x\)\(\Rightarrow\)\(\left|x-15\right|=-\left(x-15\right)=15-x\)
\(\Rightarrow15-x=15-x\)
\(\Leftrightarrow0x=0\)( Vô số các giá trị. Điều kiện: \(x< 15\))
Vậy \(x\le15\)
2) \(7x.\left(2+x\right)-7x.\left(x+3\right)=14\)
\(\Leftrightarrow7x.\left(2+x-x-3\right)=14\)
\(\Leftrightarrow-7x=14\)
\(\Leftrightarrow x=-2\)( thỏa mãn )
Vậy \(x=-2\)
Bài 2:
1) Ta có: \(A=-3x^3-2x^2+x-14\)
\(\Leftrightarrow A=-\left(3x^3+6x^2\right)+\left(4x^2+8x\right)-\left(7x+14\right)\)
\(\Leftrightarrow A=-3x^2.\left(x+2\right)+4x.\left(x+2\right)-7.\left(x+2\right)\)
\(\Leftrightarrow A=\left(x+2\right).\left(-3x^2+4x-7\right)\)
+ Thay \(x=-3\)vào biểu thức A, ta có:
\(A=\left(-3+2\right).\left(-3.9-12-7\right)\)
\(\Leftrightarrow A=\left(-1\right).\left(-46\right)\)
\(\Leftrightarrow A=46\)
Vậy \(A=46\)
2) Ta có: \(B=2xy-3x+2y\)
+ Thay \(x=-2,x=-5\)vào biểu thức B, ta có:
\(B=2.\left(-2\right).\left(-5\right)-3.\left(-2\right)+2.\left(-5\right)\)
\(\Leftrightarrow B=20+6-10\)
\(\Leftrightarrow B=16\)
Vậy \(B=16\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-2}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-2}\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=100\)
\(\Leftrightarrow x^2-5x+6-100=0\)
\(\Leftrightarrow x^2-5x-94=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{401}}{2}\\x=\dfrac{5+\sqrt{401}}{2}\end{matrix}\right.\)
7x2 - 7x( x - 3 ) = x + 2
⇔ 7x2 - 7x2 + 21x = x + 2
⇔ 21x = x + 2
⇔ 21x - x = 2
⇔ 20x = 2
⇔ x = 2/20 = 1/10