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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
\(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)
Ta đánh giá vế phải \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=\sqrt{2\left(x-4\right)^2+9}+\sqrt{3\left(x-4\right)^2+16}\ge\sqrt{9}+\sqrt{16}=3+4=7\)(Do \(\left(x-4\right)^2\ge0\forall x\))
Như vậy, để \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)(hay dấu "=" xảy ra) thì \(\left(x-4\right)^2=0\)hay x = 4
Vậy nghiệm duy nhất của phương trình là 4
f, \(\sqrt{8+\sqrt{x}}+\sqrt{5-\sqrt{x}}=5\left(đk:25\ge x\ge0\right)\)
\(< =>\sqrt{8+\sqrt{x}}-\sqrt{9}+\sqrt{5-\sqrt{x}}-\sqrt{4}=0\)
\(< =>\frac{8+\sqrt{x}-9}{\sqrt{8+\sqrt{x}}+\sqrt{9}}+\frac{5-\sqrt{x}-4}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)
\(< =>\frac{\sqrt{x}-1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{\sqrt{x}-1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)
\(< =>\left(\sqrt{x}-1\right)\left(\frac{1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}\right)=0\)
\(< =>x=1\)( dùng đk đánh giá cái ngoặc to nhé vì nó vô nghiệm )
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)
\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)
\(\Rightarrow3x+2=2\left(x+2\right)\)
\(\Rightarrow3x+2=2x+4\)
\(\Rightarrow3x-2x=4-2\)
\(\Rightarrow x=2\)
\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)
\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)
\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)
\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Rightarrow2\sqrt{x-2}=4\)
\(\Rightarrow\sqrt{x-2}=2\)
\(\Rightarrow x-2=4\)
\(\Rightarrow x=6\)
\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)
\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)
\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)
\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)
\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)
\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)
\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)
\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)
\(\Rightarrow2x^2+7x=0\)
\(\Rightarrow x\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)
\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)
\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)
\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)
\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)
\(\Rightarrow x=1\)
Nhiều vậy sao giải @@
a) Đặt \(a=\sqrt{1+x}+\sqrt{8-x}\)
\(\Leftrightarrow a^2=1+x+8-x+2\sqrt{\left(1+x\right)\left(8-x\right)}\)
\(\Leftrightarrow a^2=9+2\sqrt{\left(1+x\right)\left(8-x\right)}\)
\(\Leftrightarrow\frac{a^2-9}{2}=\sqrt{\left(1+x\right)\left(8-x\right)}\)
\(pt\Leftrightarrow a+\frac{a^2-9}{2}=3\)
\(\Leftrightarrow\frac{a^2+2a-9}{2}=3\)
\(\Leftrightarrow a^2+2a-9=6\)
\(\Leftrightarrow a^2+2a-15=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-5\end{matrix}\right.\)
Tới đây thay vào rồi tìm x
b) \(2\left(x^2+2\right)=5\sqrt{x^3+1}\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\)
Ta có : \(a^2+b^2=x^2-x+1+x+1=x^2+2\)
\(pt\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2+2b^2-5ab=0\)
\(\Leftrightarrow2a^2-4ab+2b^2-ab=0\)
\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\2a=b\end{matrix}\right.\)
Tới đây thay vào rồi lại giải tiếp
p/s: Mình bận rồi, bao giờ rảnh giải tiếp