giúp e 2 câu cuối nữa ạ :v e cảm ơn nhiều lắm ạ
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\(b,\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{\sqrt{15}}=\dfrac{\sqrt{2}}{\sqrt{5}}\)
\(d,\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\left(\sqrt{ab}-\sqrt{bc}\right)}=\sqrt{ab}+\sqrt{bc}=\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)\)
\(e,\left(a\sqrt{\dfrac{a}{b}+2\sqrt{ab}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)
\(=a\left(\sqrt{\dfrac{a}{b}+\dfrac{2b.\sqrt{ab}}{b}}+b\sqrt{\dfrac{a}{b}}\right)\sqrt{ab}\)
\(=a\sqrt{a}\sqrt{a+2b\sqrt{ab}}+b\sqrt{a^2}\)
\(=a\sqrt{a^2+2ab\sqrt{ab}}+ab\)
\(=a\left(\sqrt{a^2+2ab\sqrt{ab}}+b\right)\)
\(f,\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\left(a-\sqrt{a}+1-\sqrt{a}\right)\)
\(=\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2\)
\(=\left(a-1\right)^2=a^2-2a+1\)
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Hướng làm:
Thấy cả tử mẫu cộng lại đều bằng 2021 → Cộng thêm 1 rồi quy đồng với mỗi phân thức
\(\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\\ \Leftrightarrow\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\\ \Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}\right)=0\\ \Leftrightarrow x+2021=0\Leftrightarrow x=-2021\)
\(< =>\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\)
\(< =>\dfrac{x+2+2019}{2019}+\dfrac{x+3+2018}{2018}=\dfrac{x+4+2017}{2017}+\dfrac{x+2021}{2021}\)
\(< =>\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\)
\(< =>\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}=\right)=0\)
\(< =>x+2021=0< =>x=-2021\)
Vậy....
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a) Đặt \(a=x^2+x\)
Đa thức trở thành: \(a^2-14a+24=\left(a^2-14a+49\right)-25=\left(a-7\right)^2-25=\left(a-7-5\right)\left(a-7+5\right)=\left(a-12\right)\left(a-2\right)\)
Thay a:
\(\left(a-12\right)\left(a-2\right)=\left(x^2+x-12\right)\left(x^2+x-2\right)\)
b) Đặt \(a=x^2+x\)
Đa thức trở thành:
\(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)-12=a^2+4a-12=\left(a^2+4x+4\right)-16=\left(a+2\right)^2-16=\left(a+2-4\right)\left(a+2+4\right)=\left(a-2\right)\left(a+6\right)\)
Thay a:
\(\left(a-2\right)\left(a+6\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
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https://doc-04-90-docs.googleusercontent.com/docs/securesc/4ius91d4i6l7vmjv03uus8n8e1jouf18/nncflvovvggj7929t6q17u30r1267jdr/1638346725000/16593582377474649600/01291187093379538302Z/1HxTe2o4vk4nmppw3mBRNG3Mborq9nJWy?e=download&nonce=fkdac3vjm3rao&user=01291187093379538302Z&hash=p5ovh627ujk6q6qfgi2i0tqtv2b7jgub
Mik chỉ biết dc thế này thui
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\notin\left\{0;-9\right\}\)
Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)
Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)
\(\Leftrightarrow9x^2+81x+180=0\)
\(\Leftrightarrow x^2+9x+20=0\)
\(\Leftrightarrow x^2+4x+5x+20=0\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-4;-5}
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Câu 2:
Ta có: \(\sqrt{x^2-4x+4}=x-1\)
\(\Leftrightarrow2-x=x-1\left(x< 2\right)\)
\(\Leftrightarrow-2x=-3\)
hay \(x=\dfrac{3}{2}\left(tm\right)\)
Câu 69:
Ta có:
\(f(x)+f(y)=1\Leftrightarrow \frac{9^x}{9^x+m^2}+\frac{9^y}{9^y+m^2}=1\)
\(\Leftrightarrow \frac{9^x}{9^x+m^2}=1-\frac{9^y}{9^y+m^2}=\frac{m^2}{9^y+m^2}\)
\(\Leftrightarrow 9^{x+y}=m^4\Leftrightarrow (3^{x+y}-m^2)(3^{x+y}+m^2)=0\)
\(\Rightarrow 3^{x+y}=m^2\) (do \(3^{x+y}>0; m^2\geq 0\Rightarrow 3^{x+y}+m^2>0\) ) (1)
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Tiếp theo: \(e^{x+y}\leq e(x+y)\Leftrightarrow e^{x+y-1}\leq x+y\)
Đặt \(x+y=k\Rightarrow e^{k-1}\leq k\Leftrightarrow e^{k-1}-k\leq 0\)
Đặt \(e^{k-1}-k=f(k)\Rightarrow f(k)\leq 0(*)\)
Có: \(f'(k)=e^{k-1}-1=0\Leftrightarrow k=1\)
Lập bảng biến thiên ta thấy rằng \(f(k)_{\min}=f(1)=0\) hay \(f(k)\geq 0(**)\)
Từ \((1);(2)\Rightarrow f(k)=0\) hay \(k=1\Leftrightarrow x+y=1\)
Thay vào (1) ta có \(m^2=3\Leftrightarrow m=\pm \sqrt{3}\)
Vậy có 2 giá trị m thỏa mãn. đáp án D
Câu 70:
Để hai pt lần lượt có hai nghiệm phân biệt thì
\(\Delta _1=\Delta_2=b^2-20a>0\Leftrightarrow b^2> 20a\) (1)
Khi đó, áp dụng hệ thức Viete ta có:
Đối với PT 1: \(\ln x_1+\ln x_2=\frac{-b}{a}\Leftrightarrow \ln (x_1x_2)=\frac{-b}{a}\)
\(\Leftrightarrow x_1x_2=e^{\frac{-b}{a}}\)
Đối với PT 2: \(\log x_1+\log x_2=\frac{-b}{5}\Leftrightarrow \log (x_1x_2)=\frac{-b}{5}\)
\(\Leftrightarrow x_3x_4=10^{\frac{-b}{5}}\)
Vì \(x_1x_2> x_3x_4\Leftrightarrow e^{\frac{-b}{a}}>10^{\frac{-b}{5}}\)
\(\Leftrightarrow 10^{\frac{-b}{a\ln 10}}> 10^{\frac{-b}{5}}\)
\(\Leftrightarrow \frac{-b}{a\ln 10}>\frac{-b}{5}\Leftrightarrow a>\frac{5}{\ln 10}\)
\(\Leftrightarrow a> 2,71...\Rightarrow a\geq 3\) (vì a nguyên dương)
Theo (1) ta có: \(b^2>20a\geq 60\Rightarrow b\geq 8\) (do b nguyên dương)
Vậy \(2a+3b\geq 2.3+3.8\Leftrightarrow 2a+3b\geq 30\)
Đáp án A