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Đặt \(\sqrt[3]{x+45}=a\Rightarrow a^3=x+45\)
\(\sqrt[3]{x-16}=b\Rightarrow b^3=x-16\)
Ta có:\(\hept{\begin{cases}a-b=1\\a^3-b^3=61\end{cases}\Rightarrow\hept{\begin{cases}b=a-1\\\left(a-b\right)^3+3ab\left(a-b\right)=61\end{cases}}}\)
\(\Rightarrow1+3a\left(a-1\right)=61\) (vì a-b=1)
\(\Leftrightarrow a^2-a-20=0\)
\(\Leftrightarrow\left(a-5\right)\left(a+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=5\\a=-4\end{cases}\Rightarrow\orbr{\begin{cases}a^3=125\\a^3=-64\end{cases}\Rightarrow}\orbr{\begin{cases}x=80\\x=-109\end{cases}}}\)
Vậy nghiệm của pt là: x=80;x=-109
Giải phương trình vô tỉ
1) \(\sqrt{1-\sqrt{x}}+\sqrt{4+x}=3\)
2) \(\sqrt[3]{x+45}-\sqrt[3]{x-16}=1\)
Câu 1:
ĐK: \(0\leq x\leq 1\)
Áp dụng bđt Bunhiacopxky:
\(\text{VT}^2=(\sqrt{1-\sqrt{x}}+\sqrt{4+x})^2\leq [1-\sqrt{x}+\frac{4+x}{2}](1+2)\)
\(\Leftrightarrow \text{VT}^2\leq 3\left(3+\frac{x-2\sqrt{x}}{2}\right)\)
Vì \(0\leq x\leq 1\Rightarrow x-2\sqrt{x}\leq \sqrt{x}-2\sqrt{x}=-\sqrt{x}\leq 0\)
Do đó: \(\text{VT}^2\leq 3.3=9\Rightarrow \text{VT}\leq 3\)
Dấu bằng xảy ra khi :
\(\frac{\sqrt{1-\sqrt{x}}}{1}=\frac{\sqrt{4+x}}{2}; x=\sqrt{x}\Rightarrow x=0\)
2)
\(\sqrt[3]{x+45}-\sqrt[3]{x-16}=1\)
Đặt \(\sqrt[3]{x+45}=a; \sqrt[3]{x-16}=b\). Ta thu được HPT:
\(\left\{\begin{matrix} a-b=1\\ a^3-b^3=61\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} a-b=1\\ (a-b)^3+3ab(a-b)=61\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} a-b=1\\ 1+3ab=61\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a-b=1\\ ab=20\end{matrix}\right.\)
Thay \(a=b+1\Rightarrow (b+1)b=20\)
\(\Leftrightarrow b^2+b-20=0\Leftrightarrow (b-4)(b+5)=0\)
\(\Rightarrow \left[\begin{matrix} b=4\rightarrow x=80\\ b=-5\rightarrow x=-109\end{matrix}\right.\)
\(\sqrt{x+2\sqrt{x-1}=2}\)
\(\Leftrightarrow\sqrt{x-1+2.\sqrt{x-1}.\sqrt{1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(x-1+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x^2}=2\)
\(\Leftrightarrow x=2\)
Các câu kia lm tương tự........
a) \(\sqrt{\left(2x-1\right)^2}=3\)
⇔ \(\left|2x-1\right|=3\)
⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
ĐKXĐ : \(x\ge0\)
⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)
⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)
⇔ \(7\sqrt{x}-6\sqrt{x}=5\)
⇔ \(\sqrt{x}=5\)
⇔ \(x=25\)( tm )
c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)
ĐKXĐ : \(x\ge-5\)
⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)
⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)
⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)
⇔ \(\frac{5}{4}\sqrt{x+5}=6\)
⇔ \(\sqrt{x+5}=\frac{24}{5}\)
⇔ \(x+5=\frac{576}{25}\)
⇔ \(x=\frac{451}{25}\left(tm\right)\)
\(\Leftrightarrow\left(\sqrt{x+1}+\sqrt{x+16}\right)^2=\left(\sqrt{x+4}+\sqrt{x+9}\right)^2\)
\(\Leftrightarrow x+1+x+16+2.\sqrt{\left(x+1\right).\left(x+16\right)}=x+4+x+9+2.\sqrt{\left(x+4\right).\left(x+9\right)}\)
\(\Leftrightarrow2x+17+2.\sqrt{\left(x+1\right).\left(x+16\right)}=2x+13+2.\sqrt{\left(x+4\right).\left(x+9\right)}\)
\(\Leftrightarrow4+2.\sqrt{\left(x+1\right)\left(x+16\right)}=2.\sqrt{\left(x+4\right).\left(x+9\right)}\)
\(\Leftrightarrow2.\left(2+\sqrt{\left(x+1\right)\left(x+16\right)}\right)=2.\sqrt{\left(x+4\right).\left(x+9\right)}\)
\(\Leftrightarrow\sqrt{x^2+17x+16}+1=\sqrt{x^2+13x+36}\)
Bình phương 2 vế ta được
\(x^2+17x+16+1+2.\sqrt{x^2+17x+16}=x^2+13x+36\)
\(\Leftrightarrow2.\sqrt{x^2+17x+16}=-4x+19\)
Bình phương 2 vế ta được
\(2x^2+34x+32=16x^2-152x+361\)
\(\Leftrightarrow14x^2-186x+329=0\)
\(\Delta=\left(-186\right)^2-4.14.329=16172\)
\(x_1=\frac{186-\sqrt{16172}}{26}=2,262723898\)
\(x_2=\frac{186+\sqrt{16172}}{26}=12,04496841\)
\(\sqrt{x+1}+\sqrt{x+16}=\sqrt{x+4}+\sqrt{x+9}\)
\(\left(\sqrt{x+1}+\sqrt{x+16}\right)^2=\left(\sqrt{x+4}+\sqrt{x+9}\right)^2\)
\(x+1+x+16+2\sqrt{\left(x+1\right)\left(x+16\right)}=x+4+x+9+2\sqrt{\left(x+4\right)\left(x+9\right)}\)
\(2x+17+2\sqrt{x^2+17x+16}=2x+13+2\sqrt{x^2+13x+36}\)
\(4+2\sqrt{x^2+17x+16}=2\sqrt{x^2+13x+36}\)
\(2+\sqrt{x^2+17x+16}=\sqrt{x^2+13x+36}\)
\(\left(2+\sqrt{x^2+17x+16}\right)^2=\left(\sqrt{x^2+13x+36}\right)^2\)
\(4+x^2+17x+16+4\sqrt{x^2+17x+16}=x^2+13x+36\)
\(4\sqrt{x^2+17x+16}=-4x+16\)
\(\sqrt{x^2+17x+16}=-x+4\)
\(\hept{\begin{cases}-x+4\ge0\\x^2+17x+16=\left(-x+4\right)^2\end{cases}}\)
\(\hept{\begin{cases}-x\ge-4\\x^2+17x+16=x^2-8x+16\end{cases}}\)
\(\hept{\begin{cases}x\le4\\25x=0\end{cases}}\)
\(\hept{\begin{cases}x\le4\\x=0\end{cases}}\)
\(\Rightarrow x=0\)
1)
dat \(a=\sqrt[3]{x+1};b=\sqrt[3]{7-x}\)
ta co b=2-a
a^3+b^3=x+1+7-x=8
a^3+b^3=a^3+b^3+3ab(a+b)
ab(a+b)=0
suy ra a=0 hoac b=0 hoac a=-b
<=> x=-1; x=7
a=-b
a^3=-b^3
x+1=x+7 (vo li nen vo nghiem)
cau B tuong tu
2)
tat ca cac bai tap deu chung 1 dang do la
\(\sqrt[3]{a+m}+\sqrt[3]{b-m}\)voi m la tham so
dang nay co 2 cach
C1 lap phuong VD: \(B^3=10+3\sqrt[3]{< 5+2\sqrt{13}>< 5-2\sqrt{13}>}\left(B\right)\)
B^3=10-9B
B=1 cach nay nhanh nhung kho nhin
C2 dat an
\(a=\sqrt[3]{5+2\sqrt{13}};b=\sqrt[3]{5-2\sqrt{13}}\)
de thay B=a+b
a^3+b^3=10
ab=-3
B^3=10-9B
suy ra B=1
tuong tu giai cac cau con lai.
Bài 1:
a. Đặt \(a=\sqrt[3]{x+1}\); \(b=\sqrt[3]{7-x}\). Ta có:
\(\hept{\begin{cases}a+b=2\\a^3+b^3=8\end{cases}\Leftrightarrow a^3+\left(2-a\right)^3=8\Leftrightarrow...\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}a=0\\b=2\end{cases}}\)hoặc \(\hept{\begin{cases}a=2\\b=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=0\\\sqrt[3]{7-x}=2\end{cases}}\)hoặc \(\hept{\begin{cases}\sqrt[3]{x+1}=2\\\sqrt[3]{7-x}=0\end{cases}}\)
\(\Leftrightarrow x=-1\)hoặc \(x=7\)
a) ĐKXĐ : \(x\ge5\)
Đặt \(\sqrt{x-5}=a;\sqrt[3]{3-x}=b\)(a \(\ge0\))
Khi đó phương trình thành a + b = 2
Lại có \(b^3+a^2=-2\)
=> HPT : \(\hept{\begin{cases}a+b=2\\b^3+a^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+\left(2-b\right)^2=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b^3+b^2-4b+6=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=2-b\\\left(b+3\right)\left(b^2-2b+2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2-b\\b=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=5\\b=-3\end{cases}}\)(tm)
a = 5 => x = 30 (tm)
Vậy x = 30 là nghiệm phương trình
d) Ta có \(\sqrt{25x^2-20x+4}+\sqrt{25x^2-40x+16}=0\)
<=> \(\sqrt{\left(5x-2\right)^2}+\sqrt{\left(5x-4\right)^2}=2\)
<=> |5x - 2| + |5x - 4| = 2
Lại có |5x - 2| + |5x - 4| = |5x - 2| + |4 - 5x| \(\ge\left|5x-2+4-5x\right|=2\)
Dấu "=" xảy ra <=> \(\left(5x-2\right)\left(4-5x\right)\ge0\Leftrightarrow\frac{2}{5}\le x\le\frac{4}{5}\)
Vậy \(\frac{2}{5}\le x\le\frac{4}{5}\)là nghiệm phương trình