GPT: \(x^2-3x+3=\left(3x-\frac{4}{x}+1\right)\sqrt{x-1}\)
GPT
\(x^2-3x+3=\left(4+3x-\frac{4}{x}\right)\sqrt{x-1}\)
gpt:
\(3\left(x^2-3x+1\right)+\sqrt{3\left(x^4+x^2+1\right)}=0\)
\(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)
GPT: \(\left(x+5\right)\sqrt{x+1}+1=\left(3x +4\right)^{\frac{1}{3}}\)
gpt : \(x^2-4x+5-\frac{3x}{x^2+x+1}=\left(x-1\right)\left(1-\frac{2\sqrt{1-x}}{\sqrt{x^2+x+1}}\right)\)
gpt: \(2\sqrt{3x+7}-5\sqrt[3]{x-6}=4\)
\(\left(x^2-3x+2\right)\left(x^2-12x+32\right)\le4x^2\)
\(\left(\sqrt{x+1}-1\right)\left(\sqrt{x^2-4x+7}+1\right)=x\)
Gpt:
\(\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}=\sqrt{3}-x^2\)\(\sqrt{-4x^4y^2+16x^2y+9}-\sqrt{x^2y^2-2y^2}=2\left(x^2+\frac{1}{x^2}\right)\)\(\sqrt{-x^2+3x+4}+\sqrt{-y^2+2y+2}=\sqrt{-x^2+5x+14}\)\(\sqrt{x^2+8}-\sqrt{x^2+3}=\frac{1}{2}\left(3x-1\right)\)Bài quá dễ tự làm đi
k mình mình giải cho
Bạn nói dễ mà bạn không chịu làm thì bạn nói làm gì ???
GPT
A,\(\sqrt{X+1}+\sqrt{4-X}+\sqrt{4+3X-X^2=5}\)
B,\(\sqrt{X^2-2X+5}=2-\left(X^2-1\right)^4\)
C,\(\left(X-5\right)^3=5\sqrt[3]{2X-9}-3X+6\)
D,\(\left(2X+1\right)\sqrt{\frac{X+1}{X}}=X+2+\sqrt[3]{2X^2+X^3}\)
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chương trình giải trí do obama tài trợ:
gpt:\(\sqrt{3x^2-1}+\sqrt{x^2-x}-x\sqrt{x^2+1}=\frac{1}{2\sqrt{2}}\left(7x^2-x+4\right)\)
\(PT\Leftrightarrow7x^2-x+4-2\sqrt{2\left(3x^2-1\right)}-2\sqrt{2\left(x^2-x\right)}+2x\sqrt{2\left(x^2+1\right)}=0\)
\(\Leftrightarrow\left(3x^2-1-2\sqrt{2\left(3x^2-1\right)}+2\right)+\left(x^2-x-2\sqrt{2\left(x^2-x\right)}+2\right)+\left(2x^2+2x\sqrt{2\left(x^2+1\right)}+x^2+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{3x^2-1}-\sqrt{2}\right)^2+\left(\sqrt{x^2-x}-\sqrt{2}\right)^2+\left(\sqrt{2}x+\sqrt{x^2+1}\right)^2=0\)
Dấu = xảy ra khi x = - 1
gpt : a. \(x^2-7x=6\sqrt{x+5}-30\)
b. \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x-4}\)
a) Điều kiện $x \ge -5$. Đặt $\sqrt{x+5}=a$ thì $x=a^2-5$. Thay vào ta có $$\begin{array}{l} (a^2-5)^2-7(a^2-5)=6a-30 \\ \Leftrightarrow a^4-17a^2-6a+90=0 \Leftrightarrow (a^2+6a+10)(a-3)^2=0 \end{array}$$
Vậy $a=3 \Leftrightarrow \boxed{ x= 4}$.