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7 tháng 9 2017
do \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(\Rightarrow\sqrt{x^2+x+1}>0\forall x\)
voi dk \(x\ge-1\) ta co
\(x^2+x+1=x^2+2x+1\Rightarrow x=0\)(tm)
b,\(\sqrt{4x^2-20x+25}+2x=5\)
\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}+2x=5\)
\(\Leftrightarrow\left|2x-5\right|+2x=5\)
th1 \(2x-5\ge0\Leftrightarrow x\ge\frac{5}{2}\) ta co\(2x-5+2x=5\Leftrightarrow4x=10\Rightarrow x=2.5\left(tm\right)\)
th2 \(2x-5< 0\Leftrightarrow x< \frac{5}{2}\) \(5-2x+2x=5\Leftrightarrow5=5\)
\(\Rightarrow\) dung voi moi \(x< \frac{5}{2}\)
kl \(x\le\frac{5}{2}\)
c, \(\left|x-1\right|=4\) \(\Rightarrow\orbr{\begin{cases}x-1=4\left(x\ge1\right)\\x-1=-4\left(x< 1\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-3\left(tm\right)\end{cases}}}\)
d.\(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+16}\)
=\(\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge\sqrt{4}+\sqrt{16}=6\)
ma \(-x^2-2x+5=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)
dau = xay ra \(\Leftrightarrow x=-1\)
N
2 tháng 10 2019
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
N
2 tháng 10 2019
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
HP
22 tháng 10 2020
c, ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}-1=2\\\sqrt{2x-1}-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{2x-1}=-1\left(vn\right)\end{matrix}\right.\)
\(\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tm\right)\)
HP
22 tháng 10 2020
a, ĐKXĐ: \(x\in R\)
\(\sqrt{3x^2}=x+2\)
\(\Leftrightarrow\sqrt{3}\left|x\right|=x+2\)
TH1: \(\sqrt{3}x=x+2\)
\(\Leftrightarrow\left(\sqrt{3}-1\right)x=2\)
\(\Leftrightarrow x=\sqrt{3}+1\)
TH2: \(\sqrt{3}x=-x-2\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)x=-2\)
\(\Leftrightarrow x=1-\sqrt{3}\)
NT
19 tháng 10 2015
Mấy câu này chỉ cần tìm ĐKXĐ, chuyển vế phù hợp (có thể cần tìm thêm ĐK) rồi bình phương lên, giải bình thường nhé...chứ dài vậy...ko trả lời chi tiết được đâu bạn nhé!!(tick)
Đk: \(x\ge0\)
pt đã cho \(\Leftrightarrow6\sqrt{2x+7}-\left(\dfrac{3}{2}x+\dfrac{33}{2}\right)=2\sqrt{x}-\left(\dfrac{1}{2}x+\dfrac{3}{2}\right)\)
\(\Leftrightarrow\dfrac{36\left(2x+7\right)-\left(\dfrac{3}{2}x+\dfrac{33}{2}\right)^2}{6\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}}=\dfrac{4x-\left(\dfrac{1}{2}x+\dfrac{3}{2}\right)^2}{2\sqrt{x}+\dfrac{1}{2}x+\dfrac{3}{2}}\)
\(\Leftrightarrow\dfrac{72x+252-\dfrac{9}{4}x^2-\dfrac{99}{2}x-\dfrac{1089}{4}}{6\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}}=\dfrac{4x-\dfrac{1}{4}x^2-\dfrac{3}{2}x-\dfrac{9}{4}}{2\sqrt{x}+\dfrac{1}{2}x+\dfrac{3}{2}}\)
\(\Leftrightarrow\dfrac{-\dfrac{9}{4}x^2+\dfrac{45}{2}x-\dfrac{81}{4}}{6\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}}=\dfrac{-\dfrac{1}{4}x^2+\dfrac{5}{2}x-\dfrac{9}{4}}{2\sqrt{x}+\dfrac{1}{2}x+\dfrac{3}{2}}\)
\(\Leftrightarrow\dfrac{x^2-10x+9}{-\dfrac{4}{9}\left(6\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}\right)}=\dfrac{x^2-10x+9}{-4\left(2\sqrt{x}+\dfrac{1}{2}x+\dfrac{3}{2}\right)}\)
\(\Leftrightarrow\left(x^2-10x+9\right)\left[\dfrac{9}{4\left(6+\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}\right)}-\dfrac{1}{4\left(2\sqrt{x}+\dfrac{1}{2}x+\dfrac{3}{2}\right)}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x+9=0\\\dfrac{9}{6\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}}=\dfrac{1}{2\sqrt{x}+\dfrac{1}{2}x+\dfrac{3}{2}}\end{matrix}\right.\)
Với \(x^2-10x+9=0\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\) (nhận)
pt nhỏ thứ 2 \(\Leftrightarrow18\sqrt{x}+\dfrac{9}{2}x+\dfrac{27}{2}=6\sqrt{2x+7}+\dfrac{3}{2}x+\dfrac{33}{2}\)
\(\Leftrightarrow6\sqrt{2x+7}-18\sqrt{x}=3x-3\)
\(\Leftrightarrow2\sqrt{2x+7}-6\sqrt{x}=x-1\)
\(\Leftrightarrow\dfrac{4\left(2x+7\right)-36x}{2\sqrt{2x+7}+6\sqrt{x}}=x-1\)
\(\Leftrightarrow\dfrac{28-28x}{2\sqrt{2x+7}+6\sqrt{x}}=x-1\)
\(\Leftrightarrow\left(x-1\right)\left(1+\dfrac{28}{2\sqrt{2x+7}+6\sqrt{x}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\1+\dfrac{28}{2\sqrt{2x+7}+6\sqrt{x}}=0\left(loại\right)\end{matrix}\right.\)
Vậy pt đã cho có tập nghiệm \(S=\left\{1;9\right\}\)