\(\left(x-12\right)\div4=7\)
x-12:4=7
tìm x:
a) \(\overline{x3}+\overline{3x}=12\times11\)
b) \(4\frac{3}{4}-\left(\frac{1}{2}+x\right)\div4\frac{2}{3}=2\frac{1}{2}\)
\(\dfrac{-3}{4}x\dfrac{4}{7}+\dfrac{-3}{5}x\dfrac{3}{7}-\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}x\dfrac{5}{6}-\dfrac{7}{12}\right)x\left(\dfrac{-12}{7}\right)\)
\(1,4x\dfrac{15}{19}x\left(\dfrac{4}{5}x\dfrac{-2}{3}\right)+\dfrac{7}{15}\)
Tính nha
a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)
\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)
b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)
\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)
c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)
\(=\dfrac{-56}{95}+\dfrac{7}{15}\)
\(=\dfrac{-7}{57}\)
Tìm x
\(2x.\left(x-5\right)-x.\left(3+2x\right)=26\)
\(\left(x-7\right).\left(x-5\right)-12.\left(3x-7\right)=15\)
\(4.\left(18-5x\right)-12.\left(3x-7\right)=15.\left(2x-16\right)-6.\left(x+14\right)\)
\(\left(x-1\right).\left(x^2+x+1\right)=x^3-2x\)
Tìm x biết:
\(4\left|3x-1\right|+\left|x\right|-2\left|x-5\right|+7\left|x-3\right|=12\) \(12\)
tim xEz biet:
a)\(x^2+\left(y-\frac{1}{4}\right)^4=6\)
b)\(x+\left(\frac{-31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x=y^2\)
c)\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
cách 1:=> (x - 7)^(x+1)= (x-7)^(x+11)
TH1: x-7=0 => x=7 => 0^8=0^18 (TM)
TH2: x-7=1 => x=8 (TM)
TH3: x khác 7 và 8 => x+1=x+11 => vô lý => loại
KL: x = 7 hoặc x=8
( x-7)^( x+1) - ( x-7)^(x+11) = 0
( x-7)^( x+1) - ( x-7)^(x+1)*x^10 = 0
( x-7)^( x+1) (1-x^10) = 0
tới đây dễ òi
cách 3:\(\Leftrightarrow\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)
\(\Leftrightarrow x-7=0\)hoặc x+1=x+11(vô lí)
\(\Rightarrow x=7\)
\(12\left(x-4\right)+6\left(x-2\right)-16\left(x+3\right)=7|-4|\)
12(x - 4) + 6(x - 2) - 16(x + 3) = 7|-4|
=> 12x - 48 + 6x - 12 - 16x - 48 = 7.4
=> -2x - 108 = 28
=> -2x = 28 + 108
=> -2x = 136
=> x = 136 : (-2)
=> x = -68
\(12\left(x-4\right)+6\left(x-2\right)-16\left(x+3\right)=7.|-4|\)
\(12x-48+6x-12-16x-48=7.4\)
\(\left(12x+6x-16x\right)+\left(-48-12-48\right)=28\)
\(2x-108=28\)
\(2x=28+108\)
\(2x=136\)
\(x=136\div2=68\)
1)\(\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{-5}{4}\)
2)\(\left(5+\dfrac{4}{7}\right):x=13\)
giúp mik giải câu này với
Giải hệ pt
a) \(\left\{{}\begin{matrix}x^2+8y^2=12\\x^3+2xy^2+12y=0\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^3+y^3=1\\x^7+y^7=\left(x^4+y^4\right).1\end{matrix}\right.\)
a.
Thay số 12 từ pt trên xuống dưới:
\(x^3+2xy^2+y\left(x^2+8y^2\right)=0\)
\(\Leftrightarrow x^3+x^2y+2xy^2+8y^3=0\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2y\\x=y=0\left(ktm\right)\end{matrix}\right.\)
Thế vào pt đầu:
\(\left(-2y\right)^2+8y^2=12\Leftrightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)
b.
Thế số 1 từ pt trên xuống dưới:
\(x^7+y^7=\left(x^4+y^4\right)\left(x^3+y^3\right)\)
\(\Leftrightarrow x^4y^3+x^3y^4=0\)
\(\Leftrightarrow x^3y^3\left(x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\\y=-x\end{matrix}\right.\)
Thế vào pt đầu: \(\Rightarrow\left[{}\begin{matrix}y^3=1\\x^3=1\\x^3-x^3=1\left(vô-nghiệm\right)\end{matrix}\right.\)
Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)
\(4\cdot\left|3x-1\right|+\left|x\right|-2\cdot\left|x-5\right|+7\cdot\left|x-3\right|=12\)