Giúp tôi giải toán và làm văn

\(\hept{\begin{cases}\left(6x+5\right)^{2018}\ge0\\\left|y^2-1\right|\ge0\end{cases}}\Rightarrow\left(6x+5\right)^{2018}+\left|y^2-1\right|\ge0\)

mà \(\left(6x+5\right)^{2018}+\left|y^2-1\right|\le0\Rightarrow\left(6x+5\right)^{2018}+\left|y^2-1\right|=0\)

dấu = xảy ra khi 

\(\hept{\begin{cases}\left(6x+5\right)^{2018}=0\\\left|y^2-1\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{6}\\y=\pm1\end{cases}}\)

Vậy \(x=\frac{5}{6},y=1\) hay \(x=\frac{5}{6},y=-1\)

mu chan va tri tuyet doi deu >= 0 nen ve trai >= 0

=> 6x+5 = 0 ; y^2-1=0

\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right)\div\frac{1890}{2005}+115\)

\(A=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{11}\right)}\right)\div\frac{1890}{2005}+115\)

\(A=\left(\frac{3}{5}+\frac{-3}{5}\right)\div\frac{1890}{2005}+115\)

\(A=0\div\frac{1890}{2005}+115\)

\(A=115\)

\(\frac{37-x}{3}=\frac{x-13}{7}\)

\(\Rightarrow7\left(37-x\right)=3\left(x-13\right)\)

\(\Rightarrow259-7x=3x-39\)

\(\Rightarrow259+39=3x+7x\)

\(\Rightarrow298=10x\)

\(\Rightarrow x=\frac{298}{10}=\frac{149}{5}\)

\(\frac{37-x}{3}=\frac{x-13}{7}\)

\(\Rightarrow7\left(37-x\right)=3\left(x-13\right)\)

\(\Leftrightarrow259-7x=3x-39\)

\(\Leftrightarrow259+39=3x+7x\)

\(\Leftrightarrow298=10x\)

\(\Leftrightarrow x=29.8\)

\(\frac{37-x}{3}=\frac{x-13}{7}\)

\(\Leftrightarrow\left(37-x\right).7=\left(x-13\right).3\)

\(\Rightarrow259-7x=3x-39\)

\(10x=298\Rightarrow x=29.8\)

\(\frac{x}{5}=\frac{y}{6}\Rightarrow\frac{x^2}{25}=\frac{y^2}{36}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{25}=\frac{y^2}{36}=\frac{x^2+y^2}{61}=\frac{244}{61}=4\)

\(\frac{x^2}{25}=4\Rightarrow x^2=100\Rightarrow x=\pm10\)

\(\frac{y^2}{36}=4\Rightarrow y^2=144\Rightarrow y=\pm12\)

vậy x=10, y=12 hay x=-10, y=-12

\(x^2+3=x.\left(x-1\right)+x+3=x.\left(x-1\right)+\left(x-1\right)+4\)

=> để x²+3 chia hết cho x-1 => 4 chia hết cho x-1 => x-1 thuộc Ư(4)={1,-1,2,-2,4,-4}

=> x={2,0,3,-1,5,-3}

vì x thuộc Z => x={2,0,3,-1,5,-3}

Vì f(x₁.x₂)=f(x₁0.f(x₂) nên f(4)=f(2.2)=f(2).f(2)=5.5=25

     f(8)=f(4.2)=f(4).f(4).f(2)=25.5=125

Vậy f(8)=125

Le Tu Nhan. Có cả cặp góc xen giữa r còn j. Sử dụng trường hợp cạnh góc cạnh

Kiritokidz bn ns là ý kiến mik đúng hay khẳng định trên là đúng

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